Question Consider the region bounded by the graphs of \( y=\frac{1}{7} x^{2}, y=26 \), and \( x=10 \) in the first quadrant. What is the volume solid obtained by rotating this region about the line \( x=7 \) ?

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = \frac{1}{7} x^2 \), \( y = 26 \), and \( x = 10 \) about the line \( x = 7 \), we can use the method of cylindrical shells. ### Step 1: Identify the region of integration The curves intersect at the points where \( y = \frac{1}{7} x^2 \) and \( y = 26 \). To find the intersection points, we set: \[ \frac{1}{7} x^2 = 26 \] Multiplying both sides by 7 gives: \[ x^2 = 182 \] Taking the square root, we find: \[ x = \sqrt{182} \approx 13.5 \] However, since we are bounded by \( x = 10 \), we will only consider the region from \( x = 0 \) to \( x = 10 \). ### Step 2: Set up the volume integral The volume \( V \) of the solid of revolution using the shell method is given by: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \] In this case: - The radius of the shell is the distance from the line \( x = 7 \) to the shell at position \( x \), which is \( |7 - x| \). - The height of the shell is given by the difference between the upper curve \( y = 26 \) and the lower curve \( y = \frac{1}{7} x^2 \). Thus, the height is: \[ h = 26 - \frac{1}{7} x^2 \] ### Step 3: Define the limits of integration We will integrate from \( x = 0 \) to \( x = 10 \). ### Step 4: Write the integral The volume integral becomes: \[ V = 2\pi \int_{0}^{10} (7 - x) \left(26 - \frac{1}{7} x^2\right) \, dx \] ### Step 5: Simplify the integral Now we can simplify the integrand: \[ V = 2\pi \int_{0}^{10} \left( (7 - x)(26 - \frac{1}{7} x^2) \right) \, dx \] Expanding this gives: \[ = 2\pi \int_{0}^{10} \left( 182 - 26x - \frac{7}{7} x^2 + \frac{1}{7} x^3 \right) \, dx \] \[ = 2\pi \int_{0}^{10} \left( 182 - 26x - x^2 + \frac{1}{7} x^3 \right) \, dx \] ### Step 6: Calculate the integral Now we will calculate the integral: \[ V = 2\pi \left[ 182x - 13x^2 - \frac{1}{3} x^3 + \frac{1}{28} x^4 \right]_{0}^{10} \] Calculating at \( x = 10 \): \[ = 2\pi \left[ 182(10) - 13(10^2) - \frac{1}{3}(10^3) + \frac{1}{28}(10^4) \right] \] Calculating each term: - \( 182(10) = 1820 \) - \( 13(10^2) = 1300 \) - \( \frac{1}{3}(10^3) = \frac{1000}{3} \approx 333.33 \) - \( \frac{1}{28}(10^4) = \frac{10000}{28} \approx 357.14 \) Now substituting these values into the expression: \[ = 2\pi \left[ 1820 - 1300 - 333.33 + 357.14 \right] \] Calculating the expression inside the brackets: \[ = 2\pi \left[ 1820 - 1300 - 333.33 + 357.14 \right] = 2\pi \left[ 543.81 \right] \] ### Step 7: Final volume calculation Now we can calculate the final volume: \[ V \approx 2\pi (543.81) \approx 1087.62\pi \] Thus, the volume of the solid obtained by rotating the region about the line \( x = 7 \) is: \[ \boxed{1087.62\pi} \]