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Mertsj
Find the volume of the solid obtained by rotating the region bounded by x=0 and x = 9-y^2 about the line x=-1
|dw:1330221858790:dw|I think judging by the picture the way to go is the so-called 'washer method' the graph of x=9-y^2 intersects the line x=0 at y=3 and y=-3, so those will be our bounds of integration now for the radii...
|dw:1330222034408:dw|the outer radius will be from x=-1 to f(y), so the distance is ro=f(y)-(-1) the inner radius is from x=-1 to the x-axis, so that is 1 here is an overhead view of each washer that we will use|dw:1330222180739:dw|so the integral will be\[\pi\int_{a}^{b}r_o^2-r_i^2dy=\pi\int_{-3}^{3}(f(y)+1)^2-1^2dy\]\[=\pi\int_{-3}^{3}(10-y)^2-1dy\]
whoops, typo on the last integral (forgot to that y is squared): \[V=\pi\int_{a}^{b}r_o^2-r_i^2dy=\pi\int_{-3}^{3}(f(y)+1)^2-1^2dy\]\[V=\pi\int_{-3}^{3}(10-y^2)^2-1dy\]and since this integrand is even we can write\[V=2\pi\int_{0}^{3}(10-y^2)^2-1dy\]
Thanks Turing. You are awesome!!