RBSE Solutions Class 12 Maths Chapter 11 Application of Integral Quadrature Exercise 11.2

Question 1. Find the area between the region of parabola. \( y^2 = 2x \) and \( x^2 + y^2 = 8 \).
Answer: The given equations are a parabola and a circle: Equation of circle: \( x^2 + y^2 = 8 \) ... (i) Equation of parabola: \( y^2 = 2x \) ... (ii) To find the intersection points, substitute (ii) into (i): \( x^2 + 2x = 8 \) \( x^2 + 2x - 8 = 0 \) \( (x+4)(x-2) = 0 \) So, \( x = 2 \) or \( x = -4 \). For \( x = -4 \), \( y^2 = 2(-4) = -8 \), which means \( y \) is imaginary. For \( x = 2 \), \( y^2 = 2(2) = 4 \), so \( y = \pm 2 \). Thus, the intersection points are \( (2, 2) \) and \( (2, -2) \). The coordinate of point \( A \) on the x-axis for the parabola's limit is \( (2, 0) \). The point \( C \) is \( (2, 2) \). The x-intercept of the circle is \( (2\sqrt{2}, 0) \). The required area is shaded in the figure. It is found by integrating the difference between the upper and lower curves. Due to symmetry about the x-axis, we calculate the area in the first quadrant and multiply by 2. This helps us visualize the exact boundaries. The total area is calculated by summing the area under the parabola from \( x=0 \) to \( x=2 \) and the area under the circle from \( x=2 \) to \( x=2\sqrt{2} \). Required area \( = 2 \times (\text{Area under parabola from } x=0 \text{ to } x=2 + \text{Area under circle from } x=2 \text{ to } x=2\sqrt{2}) \) \( = 2 \left[ \int_{0}^{2} \sqrt{2x} \, dx + \int_{2}^{2\sqrt{2}} \sqrt{8-x^2} \, dx \right] \) We use the formulas: \( \int x^n dx = \frac{x^{n+1}}{n+1} \) and \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \). For the first integral: \( \int_{0}^{2} \sqrt{2x} \, dx = \sqrt{2} \int_{0}^{2} x^{1/2} \, dx = \sqrt{2} \left[ \frac{2}{3}x^{3/2} \right]_{0}^{2} \) \( = \sqrt{2} \left( \frac{2}{3}(2)^{3/2} - 0 \right) = \sqrt{2} \left( \frac{2}{3} \cdot 2\sqrt{2} \right) = \frac{8}{3} \) For the second integral, with \( a^2=8 \implies a=2\sqrt{2} \): \( \int_{2}^{2\sqrt{2}} \sqrt{8-x^2} \, dx = \left[ \frac{x}{2}\sqrt{8-x^2} + \frac{8}{2}\sin^{-1}\left(\frac{x}{2\sqrt{2}}\right) \right]_{2}^{2\sqrt{2}} \) \( = \left[ \frac{x}{2}\sqrt{8-x^2} + 4\sin^{-1}\left(\frac{x}{2\sqrt{2}}\right) \right]_{2}^{2\sqrt{2}} \) Evaluate at upper limit \( x = 2\sqrt{2} \): \( \frac{2\sqrt{2}}{2}\sqrt{8-(2\sqrt{2})^2} + 4\sin^{-1}\left(\frac{2\sqrt{2}}{2\sqrt{2}}\right) = \sqrt{2}\sqrt{8-8} + 4\sin^{-1}(1) = 0 + 4\left(\frac{\pi}{2}\right) = 2\pi \) Evaluate at lower limit \( x = 2 \): \( \frac{2}{2}\sqrt{8-2^2} + 4\sin^{-1}\left(\frac{2}{2\sqrt{2}}\right) = 1\sqrt{8-4} + 4\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \sqrt{4} + 4\left(\frac{\pi}{4}\right) = 2 + \pi \) So, \( \int_{2}^{2\sqrt{2}} \sqrt{8-x^2} \, dx = 2\pi - (2+\pi) = \pi - 2 \). Now, substitute these values back into the total area formula: Required Area \( = 2 \left[ \frac{8}{3} + (\pi - 2) \right] \) \( = \frac{16}{3} + 2\pi - 4 \) \( = \frac{16 - 12}{3} + 2\pi \) \( = \frac{4}{3} + 2\pi \) Final Area \( = \left(2\pi + \frac{4}{3}\right) \) sq. unit. In simple words: First, we find where the parabola and circle meet. Then, we split the area into two parts: one under the parabola and one under the circle. We use special math rules (integration) to add these parts together and find the total area. The result is given in square units.

🎯 Exam Tip: Always sketch the given curves to clearly identify the region whose area needs to be calculated. This helps in setting the correct limits for integration.

Question 2. Find the area of the region enclosed by parabola \( 4y = 3x^2 \) and line \( 3x - 2y + 12 = 0 \).
Answer: The equations of the given parabola and line are: Parabola: \( 4y = 3x^2 \implies y = \frac{3}{4}x^2 \) Line: \( 3x - 2y + 12 = 0 \implies 2y = 3x + 12 \implies y = \frac{3}{2}x + 6 \) To find where the line and parabola meet, we set their y-values equal: \( \frac{3}{4}x^2 = \frac{3}{2}x + 6 \) Multiply by 4 to clear fractions: \( 3x^2 = 6x + 24 \) Bring all terms to one side: \( 3x^2 - 6x - 24 = 0 \) Divide by 3: \( x^2 - 2x - 8 = 0 \) Factor the quadratic equation: \( (x-4)(x+2) = 0 \) This gives the x-coordinates of the intersection points: \( x = 4 \) and \( x = -2 \). Now, find the corresponding y-coordinates using the parabola equation \( y = \frac{3}{4}x^2 \): For \( x = 4 \): \( y = \frac{3}{4}(4^2) = \frac{3}{4}(16) = 12 \). So, the point is \( (4, 12) \). For \( x = -2 \): \( y = \frac{3}{4}(-2)^2 = \frac{3}{4}(4) = 3 \). So, the point is \( (-2, 3) \). These are the points where the line and parabola intersect, which define the boundaries of the region. A clear visual representation helps to understand the integration limits. The area of the region enclosed by the curves is found by integrating the difference between the upper curve (line) and the lower curve (parabola) from \( x = -2 \) to \( x = 4 \): Area \( = \int_{-2}^{4} \left( y_{\text{line}} - y_{\text{parabola}} \right) \, dx \) \( = \int_{-2}^{4} \left( \left(\frac{3}{2}x + 6\right) - \left(\frac{3}{4}x^2\right) \right) \, dx \) Now, we find the antiderivative of the expression: \( = \left[ \frac{3}{2}\frac{x^2}{2} + 6x - \frac{3}{4}\frac{x^3}{3} \right]_{-2}^{4} \) \( = \left[ \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 \right]_{-2}^{4} \) Evaluate the expression at the upper limit \( x = 4 \): \( \left( \frac{3}{4}(4)^2 + 6(4) - \frac{1}{4}(4)^3 \right) = \left( \frac{3}{4}(16) + 24 - \frac{1}{4}(64) \right) = (12 + 24 - 16) = 20 \) Evaluate the expression at the lower limit \( x = -2 \): \( \left( \frac{3}{4}(-2)^2 + 6(-2) - \frac{1}{4}(-2)^3 \right) = \left( \frac{3}{4}(4) - 12 - \frac{1}{4}(-8) \right) = (3 - 12 + 2) = -7 \) Subtract the lower limit value from the upper limit value: Area \( = 20 - (-7) = 27 \) sq. unit. In simple words: We first find the points where the line and the curve cross each other. Then, we use integration to find the area between these two shapes, which is like adding up tiny slices of the area. The difference in the upper and lower curve functions is integrated over the x-values of intersection to get the total area.

🎯 Exam Tip: When finding the area between two curves, always determine which function is above the other within the given interval. If the curves cross, you may need to split the integral into multiple parts.

Question 3. Find the area of the region enclosed by curve \( y = \sqrt { 4-{x}^{2} } \), \( x = \sqrt {3} y \) and x-axis.
Answer: The given curves are: 1. Curve: \( y = \sqrt { 4-{x}^{2} } \). Squaring both sides, we get \( y^2 = 4-x^2 \implies x^2 + y^2 = 4 \). This represents the upper semi-circle of a circle centered at the origin with radius 2. 2. Line: \( x = \sqrt {3} y \implies y = \frac{1}{\sqrt{3}}x \). This is a straight line passing through the origin. 3. The x-axis, which is \( y=0 \). We need to find the intersection points of the line and the circle. Substitute \( y = \frac{x}{\sqrt{3}} \) into the circle equation: \( x^2 + \left(\frac{x}{\sqrt{3}}\right)^2 = 4 \) \( x^2 + \frac{x^2}{3} = 4 \) \( \frac{4x^2}{3} = 4 \) \( 4x^2 = 12 \) \( x^2 = 3 \implies x = \pm\sqrt{3} \). Since \( y = \sqrt{4-x^2} \) implies \( y \ge 0 \), we consider \( x = \sqrt{3} \) for the intersection in the first quadrant. For \( x = \sqrt{3} \), \( y = \frac{\sqrt{3}}{\sqrt{3}} = 1 \). So, the intersection point is \( (\sqrt{3}, 1) \). The area required is bounded by the line from \( x=0 \) to \( x=\sqrt{3} \), and by the circle from \( x=\sqrt{3} \) to \( x=2 \) (the x-intercept of the circle). This is a combined area under the line and the arc of the circle. Area \( = \int_{0}^{\sqrt{3}} \frac{1}{\sqrt{3}}x \, dx + \int_{\sqrt{3}}^{2} \sqrt{4-x^2} \, dx \) First integral (area under the line): \( \int_{0}^{\sqrt{3}} \frac{1}{\sqrt{3}}x \, dx = \frac{1}{\sqrt{3}} \left[ \frac{x^2}{2} \right]_{0}^{\sqrt{3}} = \frac{1}{\sqrt{3}} \left( \frac{(\sqrt{3})^2}{2} - 0 \right) = \frac{1}{\sqrt{3}} \left( \frac{3}{2} \right) = \frac{\sqrt{3}}{2} \) Second integral (area under the circle, using \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \) with \( a=2 \)): \( \int_{\sqrt{3}}^{2} \sqrt{4-x^2} \, dx = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{\sqrt{3}}^{2} \) \( = \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{\sqrt{3}}^{2} \) Evaluate at \( x=2 \): \( \left( \frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) = (0 + 2\sin^{-1}(1)) = 2\left(\frac{\pi}{2}\right) = \pi \) Evaluate at \( x=\sqrt{3} \): \( \left( \frac{\sqrt{3}}{2}\sqrt{4-(\sqrt{3})^2} + 2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \right) = \left( \frac{\sqrt{3}}{2}\sqrt{4-3} + 2\left(\frac{\pi}{3}\right) \right) = \left( \frac{\sqrt{3}}{2} + \frac{2\pi}{3} \right) \) Subtract the lower limit value from the upper limit value: \( \int_{\sqrt{3}}^{2} \sqrt{4-x^2} \, dx = \pi - \left( \frac{\sqrt{3}}{2} + \frac{2\pi}{3} \right) = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \) Add the two integral results to get the total area: Total Area \( = \frac{\sqrt{3}}{2} + \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \) sq. unit. In simple words: We find where the line and the upper half of the circle cross. Then, we split the area into two parts: a triangle-like part under the line and a curved part under the circle. We add these two areas together using special math tools (integration) to get the final area. This is a common way to find areas of complex shapes.

🎯 Exam Tip: When dealing with square roots in equations like \( y = \sqrt{4-x^2} \), remember that it typically represents only the positive (or upper) half of the curve. Pay attention to the quadrant and axis boundaries.

Question 4. Find the area of the region in the first quadrant enclosed by circle \( x^2 + y^2 = 16 \) and line \( y = x \).
Answer: The given equations are: Circle: \( x^2 + y^2 = 16 \). This is a circle centered at the origin with radius 4. Line: \( y = x \). This is a straight line passing through the origin with a slope of 1. We need to find the area in the first quadrant. First, find the intersection point of the line and the circle. Substitute \( y=x \) into the circle equation: \( x^2 + x^2 = 16 \) \( 2x^2 = 16 \) \( x^2 = 8 \implies x = \pm\sqrt{8} = \pm 2\sqrt{2} \). Since we are interested in the first quadrant, we take \( x = 2\sqrt{2} \). The corresponding \( y \) value is also \( y = 2\sqrt{2} \). So, the intersection point is \( (2\sqrt{2}, 2\sqrt{2}) \). This point is crucial for defining the limits of our integration. The area of the region is split into two parts: 1. The area under the line \( y=x \) from \( x=0 \) to \( x=2\sqrt{2} \). 2. The area under the circle \( y=\sqrt{16-x^2} \) from \( x=2\sqrt{2} \) to \( x=4 \) (the x-intercept of the circle). Area \( = \int_{0}^{2\sqrt{2}} x \, dx + \int_{2\sqrt{2}}^{4} \sqrt{16-x^2} \, dx \) First integral (area under the line): \( \int_{0}^{2\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2\sqrt{2}} = \frac{(2\sqrt{2})^2}{2} - 0 = \frac{8}{2} = 4 \) Second integral (area under the circle, using \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \) with \( a=4 \)): \( \int_{2\sqrt{2}}^{4} \sqrt{16-x^2} \, dx = \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{2\sqrt{2}}^{4} \) \( = \left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{2\sqrt{2}}^{4} \) Evaluate at \( x=4 \): \( \left( \frac{4}{2}\sqrt{16-4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) = (0 + 8\sin^{-1}(1)) = 8\left(\frac{\pi}{2}\right) = 4\pi \) Evaluate at \( x=2\sqrt{2} \): \( \left( \frac{2\sqrt{2}}{2}\sqrt{16-(2\sqrt{2})^2} + 8\sin^{-1}\left(\frac{2\sqrt{2}}{4}\right) \right) = \left( \sqrt{2}\sqrt{16-8} + 8\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right) = \left( \sqrt{2}\sqrt{8} + 8\left(\frac{\pi}{4}\right) \right) = (4 + 2\pi) \) Subtract the lower limit value from the upper limit value: \( \int_{2\sqrt{2}}^{4} \sqrt{16-x^2} \, dx = 4\pi - (4 + 2\pi) = 2\pi - 4 \) Add the results from both integrals to get the total area: Total Area \( = 4 + (2\pi - 4) = 2\pi \) sq. unit. In simple words: We find the point where the line \( y=x \) crosses the circle. Then, we split the area into two sections: one triangular part under the line and one curved part under the circle. We add these two areas together using integration to find the total area, which represents the full enclosed region.

🎯 Exam Tip: Always make sure to integrate within the specified quadrant. If the problem specifies the first quadrant, ensure all \( x \) and \( y \) values used for limits and functions remain positive.

Question 5. Find the area of common region between parabola \( y^2 = 4x \) and \( x^2 = 4y \).
Answer: The equations of the given parabolas are: Parabola 1: \( y^2 = 4x \) ... (i) Parabola 2: \( x^2 = 4y \) ... (ii) To find the intersection points, substitute \( y = \frac{x^2}{4} \) from equation (ii) into equation (i): \( \left(\frac{x^2}{4}\right)^2 = 4x \) \( \frac{x^4}{16} = 4x \) Multiply by 16: \( x^4 = 64x \) Bring all terms to one side: \( x^4 - 64x = 0 \) Factor out \( x \): \( x(x^3 - 64) = 0 \) This gives \( x=0 \) or \( x^3 = 64 \implies x = 4 \). Now, find the corresponding y-coordinates using \( y = \frac{x^2}{4} \): For \( x=0 \): \( y = \frac{0^2}{4} = 0 \). So, the point is \( (0,0) \). For \( x=4 \): \( y = \frac{4^2}{4} = \frac{16}{4} = 4 \). So, the point is \( (4,4) \). The area enclosed by these two parabolas is found by integrating the difference between the upper curve (\( y = \sqrt{4x} \)) and the lower curve (\( y = \frac{x^2}{4} \)) from \( x=0 \) to \( x=4 \). A diagram shows how the parabolas intersect, forming a closed region. Area \( = \int_{0}^{4} \left( y_{\text{upper}} - y_{\text{lower}} \right) \, dx \) From \( y^2 = 4x \), the upper curve is \( y = \sqrt{4x} = 2\sqrt{x} \). From \( x^2 = 4y \), the lower curve is \( y = \frac{x^2}{4} \). Area \( = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) \, dx \) Now, find the antiderivative of the expression: \( = \left[ 2 \cdot \frac{2}{3}x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3} \right]_{0}^{4} \) \( = \left[ \frac{4}{3}x^{3/2} - \frac{1}{12}x^3 \right]_{0}^{4} \) Evaluate the expression at the upper limit \( x = 4 \): \( \left( \frac{4}{3}(4)^{3/2} - \frac{1}{12}(4)^3 \right) = \left( \frac{4}{3}(8) - \frac{1}{12}(64) \right) = \left( \frac{32}{3} - \frac{16}{3} \right) = \frac{16}{3} \) Evaluate the expression at the lower limit \( x = 0 \): \( \left( \frac{4}{3}(0)^{3/2} - \frac{1}{12}(0)^3 \right) = (0 - 0) = 0 \) Subtract the lower limit value from the upper limit value: Total Area \( = \frac{16}{3} - 0 = \frac{16}{3} \) sq. unit. In simple words: We find the points where the two parabolas cross each other. Then, we find the area between these two curves by subtracting the lower curve's function from the upper curve's function and integrating the result. This helps us to measure the space enclosed by both curves.

🎯 Exam Tip: When dealing with two parabolas, sketch them to clearly identify which parabola is above the other in the region of interest. This prevents errors in setting up the definite integral.

Question 6. Find the area of the region in the first quadrant enclosed by curve \( x^2 + y^2 = 1 \) and \( x + y = 1 \) equation.
Answer: The given equations are: Circle: \( x^2 + y^2 = 1 \). This is a circle centered at the origin with a radius of 1. Line: \( x + y = 1 \). This can be rewritten as \( y = 1-x \). We need to find the area in the first quadrant. First, we find the intersection points of the line and the circle. Substitute \( y = 1-x \) into the circle equation: \( x^2 + (1-x)^2 = 1 \) \( x^2 + (1 - 2x + x^2) = 1 \) \( 2x^2 - 2x + 1 = 1 \) \( 2x^2 - 2x = 0 \) Factor out \( 2x \): \( 2x(x-1) = 0 \) This gives \( x=0 \) or \( x=1 \). Now, find the corresponding y-coordinates using the line equation \( y = 1-x \): For \( x=0 \): \( y = 1-0 = 1 \). So, the point is \( (0,1) \). For \( x=1 \): \( y = 1-1 = 0 \). So, the point is \( (1,0) \). The required area is bounded by the circle arc from \( x=0 \) to \( x=1 \) and the line segment connecting \( (0,1) \) and \( (1,0) \). A visual representation makes it clear that the area is calculated by subtracting the area under the line from the area under the circle. Area \( = \int_{0}^{1} \left( y_{\text{circle}} - y_{\text{line}} \right) \, dx \) Here, \( y_{\text{circle}} = \sqrt{1-x^2} \) and \( y_{\text{line}} = 1-x \). Area \( = \int_{0}^{1} \left( \sqrt{1-x^2} - (1-x) \right) \, dx \) We can split this into two separate integrals: Area \( = \int_{0}^{1} \sqrt{1-x^2} \, dx - \int_{0}^{1} (1-x) \, dx \) First integral (area under the circle arc, using \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \) with \( a=1 \)): \( \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}\left(\frac{x}{1}\right) \right]_{0}^{1} \) Evaluate at \( x=1 \): \( \left( \frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\sin^{-1}(1) \right) = \left( 0 + \frac{1}{2}\left(\frac{\pi}{2}\right) \right) = \frac{\pi}{4} \) Evaluate at \( x=0 \): \( \left( \frac{0}{2}\sqrt{1-0^2} + \frac{1}{2}\sin^{-1}(0) \right) = (0 + 0) = 0 \) So, \( \int_{0}^{1} \sqrt{1-x^2} \, dx = \frac{\pi}{4} \) Second integral (area under the line): \( \int_{0}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} \) Evaluate at \( x=1 \): \( \left( 1 - \frac{1^2}{2} \right) = \left( 1 - \frac{1}{2} \right) = \frac{1}{2} \) Evaluate at \( x=0 \): \( \left( 0 - \frac{0^2}{2} \right) = 0 \) So, \( \int_{0}^{1} (1-x) \, dx = \frac{1}{2} \) Subtract the second integral result from the first: Total Area \( = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4} \) sq. unit. In simple words: We find the points where the line and the circle cross in the first quarter of the graph. Then, we subtract the area under the straight line from the area under the curved part of the circle. This gives us the exact size of the small region trapped between them.

🎯 Exam Tip: For regions bounded by a line and a curve that intersect the axes, visualizing the area as "area under curve minus area under line" is often the most direct approach.

Question 7. Find the area of the region enclosed by curve \( y^2 = 4ax \), and line \( y = 2a \) and y-axis.
Answer: The given equations define the boundaries of the region: Parabola: \( y^2 = 4ax \). This is a parabola opening towards the positive x-axis, with its vertex at the origin. Line 1: \( y = 2a \). This is a horizontal line. Line 2: The y-axis, which is represented by \( x=0 \). To find the intersection point of the parabola and the line \( y=2a \), substitute \( y=2a \) into the parabola equation: \( (2a)^2 = 4ax \) \( 4a^2 = 4ax \) Assuming \( a \neq 0 \), we can divide by \( 4a \): \( x = a \). So, the intersection point is \( (a, 2a) \). This point helps define the extent of the region. The region is enclosed by the y-axis, the line \( y=2a \), and the parabola \( y^2=4ax \). It is simpler to calculate this area by integrating with respect to \( y \). From the parabola equation \( y^2 = 4ax \), we can express \( x \) in terms of \( y \): \( x = \frac{y^2}{4a} \) The limits for integration with respect to \( y \) will be from \( y=0 \) (the x-axis, which is the bottom boundary of the region) to \( y=2a \) (the horizontal line). Area \( = \int_{0}^{2a} x \, dy \) Substitute the expression for \( x \): \( = \int_{0}^{2a} \frac{y^2}{4a} \, dy \) Now, find the antiderivative: \( = \frac{1}{4a} \left[ \frac{y^3}{3} \right]_{0}^{2a} \) Evaluate the expression at the upper limit \( y = 2a \): \( \frac{1}{4a} \left( \frac{(2a)^3}{3} \right) = \frac{1}{4a} \left( \frac{8a^3}{3} \right) = \frac{2a^2}{3} \) Evaluate at the lower limit \( y=0 \): \( \frac{1}{4a} \left( \frac{0^3}{3} \right) = 0 \) Subtract the lower limit value from the upper limit value: Total Area \( = \frac{2a^2}{3} - 0 = \frac{2a^2}{3} \) sq. unit. In simple words: We find where the parabola and the horizontal line meet. Instead of measuring from side to side (x-axis), we measure from bottom to top (y-axis) to find the area. We use integration with respect to y to sum up tiny horizontal strips and calculate the total enclosed space.

🎯 Exam Tip: When integrating areas bounded by a vertical line (like the y-axis), a horizontal line, and a parabola that opens left or right, it's often more efficient to integrate with respect to \( y \).

Question 8. Find the area of that portion of circle \( x^2 + y^2 = 16 \) which lies outside the parabola \( y^2 = 6x \).
Answer: The given equations are: Circle: \( x^2 + y^2 = 16 \). This is a circle centered at the origin with a radius of 4. Parabola: \( y^2 = 6x \). This is a parabola opening towards the positive x-axis. To find the intersection points, substitute \( y^2 = 6x \) into the circle equation: \( x^2 + 6x = 16 \) \( x^2 + 6x - 16 = 0 \) Factor the quadratic equation: \( (x+8)(x-2) = 0 \) This gives \( x=-8 \) or \( x=2 \). For \( x=-8 \), \( y^2 = 6(-8) = -48 \), which has no real solutions for \( y \). For \( x=2 \), \( y^2 = 6(2) = 12 \implies y = \pm\sqrt{12} = \pm 2\sqrt{3} \). So, the intersection points are \( (2, 2\sqrt{3}) \) and \( (2, -2\sqrt{3}) \). These points are crucial for defining the boundaries of integration. The total area of the circle is \( A_{\text{circle}} = \pi r^2 = \pi (4^2) = 16\pi \) sq. unit. We need to find the area inside the circle but *outside* the parabola. This can be calculated by subtracting the area common to both the circle and the parabola from the total area of the circle. The common area (inside both) is symmetric about the x-axis. The area common to both is calculated by integrating the upper curve's y-value from the parabola and then the circle. Area common \( A_{\text{common}} = 2 \left[ \int_{0}^{2} \sqrt{6x} \, dx + \int_{2}^{4} \sqrt{16-x^2} \, dx \right] \) First integral (area under parabola \( y=\sqrt{6x} \)): \( \int_{0}^{2} \sqrt{6x} \, dx = \sqrt{6} \int_{0}^{2} x^{1/2} \, dx = \sqrt{6} \left[ \frac{2}{3}x^{3/2} \right]_{0}^{2} \) \( = \sqrt{6} \left( \frac{2}{3}(2)^{3/2} - 0 \right) = \sqrt{6} \left( \frac{2}{3} \cdot 2\sqrt{2} \right) = \frac{4\sqrt{12}}{3} = \frac{4 \cdot 2\sqrt{3}}{3} = \frac{8\sqrt{3}}{3} \) Second integral (area under circle \( y=\sqrt{16-x^2} \), using \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \) with \( a=4 \)): \( \int_{2}^{4} \sqrt{16-x^2} \, dx = \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{2}^{4} \) \( = \left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{2}^{4} \) Evaluate at \( x=4 \): \( \left( \frac{4}{2}\sqrt{16-4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) = (0 + 8\sin^{-1}(1)) = 8\left(\frac{\pi}{2}\right) = 4\pi \) Evaluate at \( x=2 \): \( \left( \frac{2}{2}\sqrt{16-2^2} + 8\sin^{-1}\left(\frac{2}{4}\right) \right) = \left( 1\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right) \right) = \left( 2\sqrt{3} + 8\left(\frac{\pi}{6}\right) \right) = 2\sqrt{3} + \frac{4\pi}{3} \) Subtract the lower limit value from the upper limit value: \( \int_{2}^{4} \sqrt{16-x^2} \, dx = 4\pi - \left( 2\sqrt{3} + \frac{4\pi}{3} \right) = 4\pi - 2\sqrt{3} - \frac{4\pi}{3} = \frac{12\pi - 4\pi}{3} - 2\sqrt{3} = \frac{8\pi}{3} - 2\sqrt{3} \) Now, sum the parts for \( A_{\text{common}} \): \( A_{\text{common}} = 2 \left[ \frac{8\sqrt{3}}{3} + \left( \frac{8\pi}{3} - 2\sqrt{3} \right) \right] \) \( = 2 \left[ \frac{8\sqrt{3}}{3} + \frac{8\pi}{3} - \frac{6\sqrt{3}}{3} \right] \) \( = 2 \left[ \frac{2\sqrt{3} + 8\pi}{3} \right] = \frac{4\sqrt{3} + 16\pi}{3} \) Finally, subtract this common area from the total circle area: Required Area \( = A_{\text{circle}} - A_{\text{common}} \) \( = 16\pi - \left( \frac{16\pi + 4\sqrt{3}}{3} \right) \) \( = \frac{48\pi - 16\pi - 4\sqrt{3}}{3} = \frac{32\pi - 4\sqrt{3}}{3} \) sq. unit. In simple words: We find where the circle and parabola cross. Then, we find the total area of the circle. We also calculate the area that is shared by both the circle and the parabola using integration. Finally, we subtract this shared area from the total circle area to find the part of the circle that is outside the parabola.

🎯 Exam Tip: When finding areas 'outside' one curve but 'inside' another, subtract the overlapping area from the larger enclosing area. Careful visualization or sketching is key to setting up the correct integrals and limits.

Question 9. By using integrating method, find the area of ∆ABC whose coordinates of vertices are A(2, 0), B(4, 5), C(6, 3).
Answer: The coordinates of the vertices of the triangle are \( A(2, 0) \), \( B(4, 5) \), and \( C(6, 3) \). To find the area using integration, we need to determine the equations of the lines forming the sides of the triangle. 1. **Equation of line AB** (passing through \( A(2,0) \) and \( B(4,5) \)): Using the two-point form \( \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \): \( \frac{y-0}{x-2} = \frac{5-0}{4-2} = \frac{5}{2} \) \( y_{\text{AB}} = \frac{5}{2}(x-2) \) 2. **Equation of line BC** (passing through \( B(4,5) \) and \( C(6,3) \)): \( \frac{y-5}{x-4} = \frac{3-5}{6-4} = \frac{-2}{2} = -1 \) \( y-5 = -1(x-4) \implies y_{\text{BC}} = -x + 4 + 5 \implies y_{\text{BC}} = -x + 9 \) 3. **Equation of line AC** (passing through \( A(2,0) \) and \( C(6,3) \)): \( \frac{y-0}{x-2} = \frac{3-0}{6-2} = \frac{3}{4} \) \( y_{\text{AC}} = \frac{3}{4}(x-2) \) The area of the triangle can be calculated by summing the areas of trapezoids formed by the lines AB and BC with the x-axis, and subtracting the area of the trapezoid formed by line AC with the x-axis. This method involves integrating the functions with respect to \( x \). Area of \( \triangle ABC = \int_{2}^{4} y_{\text{AB}} \, dx + \int_{4}^{6} y_{\text{BC}} \, dx - \int_{2}^{6} y_{\text{AC}} \, dx \) Now, we calculate each integral: 1. **Integral for AB**: \( \int_{2}^{4} \frac{5}{2}(x-2) \, dx = \frac{5}{2} \left[ \frac{x^2}{2} - 2x \right]_{2}^{4} \) \( = \frac{5}{2} \left[ \left(\frac{4^2}{2} - 2(4)\right) - \left(\frac{2^2}{2} - 2(2)\right) \right] \) \( = \frac{5}{2} \left[ (8 - 8) - (2 - 4) \right] = \frac{5}{2} [0 - (-2)] = \frac{5}{2}(2) = 5 \) 2. **Integral for BC**: \( \int_{4}^{6} (-x+9) \, dx = \left[ -\frac{x^2}{2} + 9x \right]_{4}^{6} \) \( = \left[ \left(-\frac{6^2}{2} + 9(6)\right) - \left(-\frac{4^2}{2} + 9(4)\right) \right] \) \( = \left[ (-18 + 54) - (-8 + 36) \right] = [36 - 28] = 8 \) 3. **Integral for AC**: \( \int_{2}^{6} \frac{3}{4}(x-2) \, dx = \frac{3}{4} \left[ \frac{x^2}{2} - 2x \right]_{2}^{6} \) \( = \frac{3}{4} \left[ \left(\frac{6^2}{2} - 2(6)\right) - \left(\frac{2^2}{2} - 2(2)\right) \right] \) \( = \frac{3}{4} \left[ (18 - 12) - (2 - 4) \right] = \frac{3}{4} [6 - (-2)] = \frac{3}{4}(8) = 6 \) Finally, combine the results: Total Area \( = 5 + 8 - 6 = 7 \) sq. unit. In simple words: To find the area of the triangle, we first find the equations for each of its three sides. Then, we add up the areas under the top two sides and subtract the area under the bottom side using integration. This method breaks down the triangle into simpler shapes, allowing us to calculate its area precisely.

🎯 Exam Tip: Remember that the area of a polygon can be found by summing signed areas of trapezoids formed by its sides with the x-axis. Always order the vertices correctly or use the Shoelace formula for verification.

Question 10. By using intersection method, find the area of triangular prism equations of whose sides are \( 3x - 2y + 3 = 0 \), \( x + 2y - 7 = 0 \) and \( x - 2y + 1 = 0 \).
Answer: The equations of the sides of the triangular region are given as: Line 1 (\( L_1 \)): \( 3x - 2y + 3 = 0 \implies y = \frac{3}{2}x + \frac{3}{2} \) Line 2 (\( L_2 \)): \( x + 2y - 7 = 0 \implies y = -\frac{1}{2}x + \frac{7}{2} \) Line 3 (\( L_3 \)): \( x - 2y + 1 = 0 \implies y = \frac{1}{2}x + \frac{1}{2} \) To find the vertices of the triangle, we find the intersection points of these lines: 1. **Intersection of \( L_1 \) and \( L_2 \)** (Vertex A): Set the y-values equal: \( \frac{3}{2}x + \frac{3}{2} = -\frac{1}{2}x + \frac{7}{2} \) Multiply by 2: \( 3x + 3 = -x + 7 \) \( 4x = 4 \implies x=1 \) Substitute \( x=1 \) into \( y = \frac{3}{2}x + \frac{3}{2} \): \( y = \frac{3}{2}(1) + \frac{3}{2} = \frac{6}{2} = 3 \). So, Vertex A is \( (1,3) \). 2. **Intersection of \( L_2 \) and \( L_3 \)** (Vertex C): Set the y-values equal: \( -\frac{1}{2}x + \frac{7}{2} = \frac{1}{2}x + \frac{1}{2} \) Multiply by 2: \( -x + 7 = x + 1 \) \( 6 = 2x \implies x=3 \) Substitute \( x=3 \) into \( y = \frac{1}{2}x + \frac{1}{2} \): \( y = \frac{1}{2}(3) + \frac{1}{2} = \frac{4}{2} = 2 \). So, Vertex C is \( (3,2) \). 3. **Intersection of \( L_3 \) and \( L_1 \)** (Vertex B): Set the y-values equal: \( \frac{1}{2}x + \frac{1}{2} = \frac{3}{2}x + \frac{3}{2} \) Multiply by 2: \( x + 1 = 3x + 3 \) \( -2 = 2x \implies x=-1 \) Substitute \( x=-1 \) into \( y = \frac{1}{2}x + \frac{1}{2} \): \( y = \frac{1}{2}(-1) + \frac{1}{2} = 0 \). So, Vertex B is \( (-1,0) \). The vertices of the triangle are \( A(1,3) \), \( B(-1,0) \), and \( C(3,2) \). To find the area using integration, we sum the areas under the upper lines and subtract the area under the lower line with respect to the x-axis. A visual sketch of the graph will help understand the limits of integration for each segment. Area \( = \int_{-1}^{1} y_{\text{L1}} \, dx + \int_{1}^{3} y_{\text{L2}} \, dx - \int_{-1}^{3} y_{\text{L3}} \, dx \) Here, \( y_{\text{L1}} = \frac{3}{2}x + \frac{3}{2} \), \( y_{\text{L2}} = -\frac{1}{2}x + \frac{7}{2} \), and \( y_{\text{L3}} = \frac{1}{2}x + \frac{1}{2} \). 1. **Integral for Line 1 (from \( x=-1 \) to \( x=1 \))**: \( \int_{-1}^{1} \left(\frac{3}{2}x + \frac{3}{2}\right) \, dx = \left[ \frac{3}{4}x^2 + \frac{3}{2}x \right]_{-1}^{1} \) \( = \left( \frac{3}{4}(1)^2 + \frac{3}{2}(1) \right) - \left( \frac{3}{4}(-1)^2 + \frac{3}{2}(-1) \right) \) \( = \left( \frac{3}{4} + \frac{6}{4} \right) - \left( \frac{3}{4} - \frac{6}{4} \right) = \frac{9}{4} - \left(-\frac{3}{4}\right) = \frac{12}{4} = 3 \) 2. **Integral for Line 2 (from \( x=1 \) to \( x=3 \))**: \( \int_{1}^{3} \left(-\frac{1}{2}x + \frac{7}{2}\right) \, dx = \left[ -\frac{1}{4}x^2 + \frac{7}{2}x \right]_{1}^{3} \) \( = \left( -\frac{1}{4}(3)^2 + \frac{7}{2}(3) \right) - \left( -\frac{1}{4}(1)^2 + \frac{7}{2}(1) \right) \) \( = \left( -\frac{9}{4} + \frac{21}{2} \right) - \left( -\frac{1}{4} + \frac{7}{2} \right) = \left( \frac{-9+42}{4} \right) - \left( \frac{-1+14}{4} \right) = \frac{33}{4} - \frac{13}{4} = \frac{20}{4} = 5 \) 3. **Integral for Line 3 (from \( x=-1 \) to \( x=3 \))**: \( \int_{-1}^{3} \left(\frac{1}{2}x + \frac{1}{2}\right) \, dx = \left[ \frac{1}{4}x^2 + \frac{1}{2}x \right]_{-1}^{3} \) \( = \left( \frac{1}{4}(3)^2 + \frac{1}{2}(3) \right) - \left( \frac{1}{4}(-1)^2 + \frac{1}{2}(-1) \right) \) \( = \left( \frac{9}{4} + \frac{6}{4} \right) - \left( \frac{1}{4} - \frac{2}{4} \right) = \frac{15}{4} - \left(-\frac{1}{4}\right) = \frac{16}{4} = 4 \) Combine the results: Total Area \( = 3 + 5 - 4 = 4 \) sq. unit. In simple words: First, we find the three corner points (vertices) where the given lines cross each other. Then, we use a method of integration where we sum the areas under the upper boundary lines and subtract the area under the lowest boundary line. This helps us calculate the exact size of the triangular region.

🎯 Exam Tip: When using the integration method for a triangle, correctly identifying the upper and lower boundary lines within each \( x \)-interval is essential. A precise graph helps avoid errors in setting up the integrals.