Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.9

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1. Find by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = 2x², y = 0 and x = 1
Answer: To find the volume, we revolve the region defined by \( y = 2x^2 \), \( y = 0 \), and \( x = 1 \) around the x-axis. The solid formed by this rotation is obtained by summing up small disks. The volume \( V \) is given by the integral: \[ V = \pi \int_0^1 y^2 \,dx \] Substitute \( y = 2x^2 \): \[ V = \pi \int_0^1 (2x^2)^2 \,dx \] \[ V = \pi \int_0^1 4x^4 \,dx \] Now, integrate \( 4x^4 \): \[ V = \pi \left[ \frac{4x^5}{5} \right]_0^1 \] Apply the limits of integration: \[ V = \pi \left( \frac{4(1)^5}{5} - \frac{4(0)^5}{5} \right) \] \[ V = \pi \left( \frac{4}{5} - 0 \right) \] \[ V = \frac{4\pi}{5} \] So, the required volume is \( \frac{4\pi}{5} \) cubic units. This method is often called the disk method, which works well when rotating a region about an axis to form a solid.
In simple words: We find the volume of a shape created by spinning a curve around the x-axis. We use a formula that involves pi and an integral, plugging in the curve's equation and calculating the answer.

🎯 Exam Tip: Remember to square the function \( y \) before integrating when using the disk method for revolving around the x-axis, and always check your limits of integration.

Question 2. Find, by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = e\(^{-2x}\), y = 0, x = 0 and x = 1.
Answer: We need to calculate the volume of the solid formed by rotating the region bounded by \( y = e^{-2x} \), \( y = 0 \), \( x = 0 \), and \( x = 1 \) around the x-axis. We will use the disk method for this. The formula for the volume \( V \) is: \[ V = \pi \int_0^1 y^2 \,dx \] Substitute \( y = e^{-2x} \): \[ V = \pi \int_0^1 (e^{-2x})^2 \,dx \] \[ V = \pi \int_0^1 e^{-4x} \,dx \] Now, integrate \( e^{-4x} \). Remember that the integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} \): \[ V = \pi \left[ \frac{e^{-4x}}{-4} \right]_0^1 \] Apply the limits of integration from 0 to 1: \[ V = \pi \left( \frac{e^{-4(1)}}{-4} - \frac{e^{-4(0)}}{-4} \right) \] \[ V = \pi \left( \frac{e^{-4}}{-4} - \frac{e^0}{-4} \right) \] Since \( e^0 = 1 \): \[ V = \pi \left( -\frac{e^{-4}}{4} + \frac{1}{4} \right) \] \[ V = \frac{\pi}{4} (1 - e^{-4}) \] So, the required volume is \( \frac{\pi}{4} (1 - e^{-4}) \) cubic units. This problem uses the fundamental theorem of calculus to evaluate the definite integral.
In simple words: We spin the area under the curve \( y = e^{-2x} \) around the x-axis to make a 3D shape. We use a special formula with integration to find how much space this shape takes up.

🎯 Exam Tip: Be careful with the negative exponent during integration; ensure you divide by the derivative of the exponent (in this case, -4) and correctly apply the limits, especially for \( e^0 \).

Question 3. Find, by integration, the volume of the solid generated by revolving about the y axis, the region enclosed by x² = 1 + y and y = 3.
Answer: We need to find the volume of the solid formed by rotating the region bounded by \( x^2 = 1 + y \) and \( y = 3 \) around the y-axis. When revolving around the y-axis, we use the formula involving \( x^2 \) and integrate with respect to \( y \). First, find the lower limit for \( y \). When \( x=0 \), \( 0 = 1+y \implies y = -1 \). So, the limits for \( y \) are from -1 to 3. The formula for the volume \( V \) is: \[ V = \pi \int_a^b x^2 \,dy \] Substitute \( x^2 = 1 + y \): \[ V = \pi \int_{-1}^3 (1 + y) \,dy \] Now, integrate \( 1 + y \): \[ V = \pi \left[ y + \frac{y^2}{2} \right]_{-1}^3 \] Apply the limits of integration from -1 to 3: \[ V = \pi \left[ \left( 3 + \frac{3^2}{2} \right) - \left( (-1) + \frac{(-1)^2}{2} \right) \right] \] \[ V = \pi \left[ \left( 3 + \frac{9}{2} \right) - \left( -1 + \frac{1}{2} \right) \right] \] Simplify the terms inside the brackets: \[ V = \pi \left[ \left( \frac{6}{2} + \frac{9}{2} \right) - \left( -\frac{2}{2} + \frac{1}{2} \right) \right] \] \[ V = \pi \left[ \frac{15}{2} - \left( -\frac{1}{2} \right) \right] \] \[ V = \pi \left[ \frac{15}{2} + \frac{1}{2} \right] \] \[ V = \pi \left[ \frac{16}{2} \right] \] \[ V = \pi (8) \] \[ V = 8\pi \] So, the required volume is \( 8\pi \) cubic units. When revolving around the y-axis, it's often more convenient to express the radius in terms of y.
In simple words: We spin the area defined by a curve and a line around the y-axis. To find the volume of this 3D shape, we use a formula involving pi and integrate with respect to y, then put in the correct start and end points.

🎯 Exam Tip: For revolution about the y-axis, ensure your function is expressed as \( x^2 \) in terms of \( y \) and your integration limits are for \( y \)-values. This is crucial for using the disk or washer method correctly.

Question 4. The region enclosed between the graphs of y = x and y = x² is denoted by R. Find the volume generated when R is rotated through 360° about x axis.
Answer: First, we need to find where the two curves \( y = x \) and \( y = x^2 \) intersect. Set the equations equal to each other: \[ x^2 = x \] Move \( x \) to one side: \[ x^2 - x = 0 \] Factor out \( x \): \[ x(x - 1) = 0 \] This gives two intersection points:
\( x = 0 \)
\( x = 1 \) When \( x = 0 \), \( y = 0 \) (from \( y = x \)). So, the point is \( (0, 0) \). When \( x = 1 \), \( y = 1 \) (from \( y = x \)). So, the point is \( (1, 1) \). The region R is between \( y = x \) and \( y = x^2 \) from \( x=0 \) to \( x=1 \). Since \( y=x \) is above \( y=x^2 \) in this interval (e.g., at \( x=0.5 \), \( y=0.5 \) and \( y=0.25 \)), we use the washer method to find the volume generated by revolving R about the x-axis. The washer method is used when there's a hollow space. The outer radius \( R(x) \) is \( y = x \), and the inner radius \( r(x) \) is \( y = x^2 \). The volume \( V \) is given by: \[ V = \pi \int_0^1 (R(x)^2 - r(x)^2) \,dx \] \[ V = \pi \int_0^1 (x^2 - (x^2)^2) \,dx \] \[ V = \pi \int_0^1 (x^2 - x^4) \,dx \] Now, integrate term by term: \[ V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 \] Apply the limits of integration from 0 to 1: \[ V = \pi \left[ \left( \frac{1^3}{3} - \frac{1^5}{5} \right) - \left( \frac{0^3}{3} - \frac{0^5}{5} \right) \right] \] \[ V = \pi \left[ \left( \frac{1}{3} - \frac{1}{5} \right) - (0 - 0) \right] \] Find a common denominator for the fractions: \[ V = \pi \left[ \frac{5}{15} - \frac{3}{15} \right] \] \[ V = \pi \left[ \frac{2}{15} \right] \] \[ V = \frac{2\pi}{15} \] So, the required volume is \( \frac{2\pi}{15} \) cubic units. The washer method is essential for regions between two curves.
In simple words: We find the area between two curves, then spin this area around the x-axis to make a 3D shape. We use a formula that subtracts the volume of the inner shape from the outer shape to get the total volume.

🎯 Exam Tip: When using the washer method, always correctly identify which function represents the outer radius and which represents the inner radius to avoid sign errors in the integration setup.

Question 5. Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown to figure
Answer: To find the volume of a frustum using integration, we first need to find the equation of the line that, when revolved around the x-axis, forms the frustum. The frustum is a part of a cone. We are given the dimensions, which imply two points on the line. Let's assume the frustum is generated by revolving a line segment from \( (0,1) \) to \( (2,2) \) around the x-axis. The equation of a straight line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \). Using points A\( (0, 1) \) and B\( (2, 2) \): \[ \frac{y - 1}{2 - 1} = \frac{x - 0}{2 - 0} \] \[ \frac{y - 1}{1} = \frac{x}{2} \] \[ y - 1 = \frac{x}{2} \] Solve for \( y \): \[ y = \frac{x}{2} + 1 \] \[ y = \frac{x + 2}{2} \] Now, we find the volume of the solid generated by revolving this line segment around the x-axis from \( x=0 \) to \( x=2 \). We use the disk method. The volume \( V \) is given by: \[ V = \pi \int_0^2 y^2 \,dx \] Substitute \( y = \frac{x+2}{2} \): \[ V = \pi \int_0^2 \left( \frac{x+2}{2} \right)^2 \,dx \] \[ V = \pi \int_0^2 \frac{(x+2)^2}{4} \,dx \] We can take \( \frac{1}{4} \) outside the integral: \[ V = \frac{\pi}{4} \int_0^2 (x+2)^2 \,dx \] Integrate \( (x+2)^2 \). We can use a substitution \( u = x+2 \), \( du = dx \), or simply integrate directly: \[ V = \frac{\pi}{4} \left[ \frac{(x+2)^3}{3} \right]_0^2 \] Apply the limits of integration from 0 to 2: \[ V = \frac{\pi}{4} \left[ \frac{(2+2)^3}{3} - \frac{(0+2)^3}{3} \right] \] \[ V = \frac{\pi}{4} \left[ \frac{4^3}{3} - \frac{2^3}{3} \right] \] \[ V = \frac{\pi}{4} \left[ \frac{64}{3} - \frac{8}{3} \right] \] \[ V = \frac{\pi}{4} \left[ \frac{56}{3} \right] \] Multiply the terms: \[ V = \frac{56\pi}{12} \] Simplify the fraction by dividing by 4: \[ V = \frac{14\pi}{3} \] So, the volume of the frustum is \( \frac{14\pi}{3} \) cubic meters. This calculation demonstrates how to use integration to find volumes of solids with changing radii, like a frustum.
In simple words: To find the volume of a frustum (a cone with its top cut off), we first find the equation of the slanted line that makes up its side. Then, we spin this line around the x-axis and use a math method called integration to calculate the total volume.

🎯 Exam Tip: When calculating the volume of a frustum by revolving a line, correctly determine the coordinates of the endpoints of the line segment that forms the profile, and then accurately derive the equation of that line.

Question 6. A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume using integration.
Answer: An ellipsoid is formed by revolving an ellipse around one of its axes. We are given the major axis is 20 cm and the minor axis is 10 cm. Let \( 2a \) be the length of the major axis, so \( 2a = 20 \) cm, which means \( a = 10 \) cm. Let \( 2b \) be the length of the minor axis, so \( 2b = 10 \) cm, which means \( b = 5 \) cm. Since the ellipse is revolved about its major axis (the x-axis in standard form), the equation of the ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Substitute \( a=10 \) and \( b=5 \): \[ \frac{x^2}{10^2} + \frac{y^2}{5^2} = 1 \] \[ \frac{x^2}{100} + \frac{y^2}{25} = 1 \] We need to express \( y^2 \) in terms of \( x \) to integrate with respect to \( x \): \[ \frac{y^2}{25} = 1 - \frac{x^2}{100} \] \[ y^2 = 25 \left( 1 - \frac{x^2}{100} \right) \] \[ y^2 = 25 \left( \frac{100 - x^2}{100} \right) \] \[ y^2 = \frac{1}{4} (100 - x^2) \] The revolution is about the major axis, so \( x \) ranges from \( -a \) to \( a \), which is from \( -10 \) to \( 10 \). Due to symmetry, we can integrate from \( 0 \) to \( 10 \) and multiply by 2. This helps simplify calculations. The volume \( V \) of the ellipsoid is given by: \[ V = \pi \int_{-10}^{10} y^2 \,dx \] \[ V = 2\pi \int_0^{10} y^2 \,dx \] Substitute the expression for \( y^2 \): \[ V = 2\pi \int_0^{10} \frac{1}{4} (100 - x^2) \,dx \] Take \( \frac{1}{4} \) outside the integral: \[ V = \frac{2\pi}{4} \int_0^{10} (100 - x^2) \,dx \] \[ V = \frac{\pi}{2} \int_0^{10} (100 - x^2) \,dx \] Now, integrate term by term: \[ V = \frac{\pi}{2} \left[ 100x - \frac{x^3}{3} \right]_0^{10} \] Apply the limits of integration from 0 to 10: \[ V = \frac{\pi}{2} \left[ \left( 100(10) - \frac{10^3}{3} \right) - \left( 100(0) - \frac{0^3}{3} \right) \right] \] \[ V = \frac{\pi}{2} \left[ \left( 1000 - \frac{1000}{3} \right) - (0 - 0) \right] \] Find a common denominator: \[ V = \frac{\pi}{2} \left[ \frac{3000}{3} - \frac{1000}{3} \right] \] \[ V = \frac{\pi}{2} \left[ \frac{2000}{3} \right] \] Multiply the terms: \[ V = \frac{2000\pi}{6} \] Simplify the fraction by dividing by 2: \[ V = \frac{1000\pi}{3} \] So, the volume of the watermelon (ellipsoid) is \( \frac{1000\pi}{3} \) cubic cm. The volume of an ellipsoid formed by revolving an ellipse is a classic application of integration in geometry.
In simple words: To find the volume of an ellipsoid, like a watermelon, we start with the equation of an ellipse and spin it around its longest line (major axis). We use a special integration formula to calculate the total space it fills up.

🎯 Exam Tip: When calculating the volume of a solid of revolution, correctly identifying the axis of revolution and expressing the radius in terms of the integration variable (x or y) is the most critical first step. Remember that if the region is symmetric, you can integrate from 0 to 'a' and multiply by 2 to simplify calculations.