Volumes of Revolution

Question

If x ≥ 0, find the volume of the solid obtained by rotating the region enclosed by the graphs about the line y = —5.

у = 90 - х. y = x², x= 0

(Use symbolic notation and fractions where needed.).

Answer 1

First, we must find the bounds of integration. We do this by setting the given equations y=90-x and y=x^2 equal to each other and solving for x. This gets us two answers x=9 and x=-10; however, since the question specifies the restriction that x must be greater than or equal to 0, we discard the negative solution. Our lower bound of integration is 0 since the region is bounded at the bottom by x=0 as given in the question.

Next, we must find the solid formed by rotating this bounded region around a given axis of rotation. Here, the question says that the axis of rotation is y=-5. Since this is not the y-axis, we must compute the distance from each bound to the axis of rotation and use the washer method to find the volume. Define R(x) as the outer function to be the distance from y=-5 to y=90-x and define r(x) as the inner function to be the distance from y=-5 to y=x^2. To find the distance, remember to simply subtract y from -5. Using the formula for the washer method, π∫([(R^(x)]²-[r(x)]²)dx you should be able to now evaluate this integral from the given bounds above.

Answer 2

90 - x = x²; x = - 10 or x = 9.

By washer method volume v = π∫_-10⁹[(95 - x)² - (x² + 5)²]dx = 67904π/5