Question

Find the volume of the solid generated by revolving the region bounded by
y= -x^2 +2x y=0 about the line y=2

Answer 1

A (x) = π ( outer radius )^2 - π ( inner radius )^2
A (x) = π ( 6 - 2x - x^2 - 2 )^2 - π ( x + 6 - 2 )^2
A (x) = π ( 4 - 2x - x^2 )^2 - π ( x + 4 )^2
A (x) = π ( x^4 + 4x^3 - 4x^2 - 16x + 16 ) - (x^2 + 8x + 16 )
A (x) = π ( x^4 + 4x^3 - 4x^2 - 16x + 16 - x^2 - 8x - 16 )
A (x) = π ( x^4 + 4x^3 - 5x^2 - 24x )

Answer 2

but for volume?

Answer 3

This is clearly a situation to use the washer method to calculate the volume. If you revolved the region around y = 2 and took a cross-section, it would look something like this: We also know that y = -x^2 + 2x is bounded by y = 0 so the region we are revolving is the part of the graph that has x-intercepts of 0 and 2. So we integrate the area of the cross section from 0 to 2 with respect to x:\[\int\limits_{0}^{2}(2^2)\pi - (2-(-x^2+2x))^2 \pi dx= \pi \int\limits_{0}^{2}4-(x^2-2x+2)^2dx\]\[=\pi \int\limits_{0}^{2}4-(x^2-2x+2)^2dx \approx 13.404\] I used a graphing calculator for the answer but you could do it by hand and integrate manually.

@kimmy0394