I'm trying to use the washer method to find volume of solid formed by revolving region bound by $f(x) = 2 - x^2$ and $g(x)= 1$ about the line $y=3$. So far I have $$V=\int_{-1}^1 \pi(x^4-4x^2+3)dx$$
I don't know if this is correct and am having trouble if anyone can verify the answer it would be great help. Thank you, -Henry
The outer radius $R$ for our washer is always equal to 2. The inner radius $r$ is given by $$r=3-(2-x^2)$$
So the volume $v$ of one slice is
$$v=\pi(R^2-r^2)$$ $$=\pi(4-(x^2+1)^2)$$ $$=\pi(3-2x^2-x^4)$$
Now to find the total volume $V$ we integrate the washers from -1 to 1.
$$V=\int_{-1}^1\pi(3-2x^2-x^4)dx$$ $$=2\pi\left[3x-\frac 2 3 x^3-\frac 1 5 x^5\right]_0^1$$ $$=\frac{64\pi}{15}$$