Volume of $2$ Solids of Revolution

Question

The region bounded by $ x = y^{2}$ and $x = 2-y^{2}$, revolved around the line $x = 3$.

By using the Washer method, I acquire: $$\pi \int_{-1}^{1} (y^2)^2 - (1-y^2)^2 \space dy$$

However after integration, the final result does not match those of an online calculator. I'm just wondering where I went wrong with the equation. The concept of revolution is a bit confusing to me, especially in this question where the functions are symmetrical to each other, and I'm not sure which is the top/bottom or right/left function.

I would appreciate any help.

Answer 1

It should be:$V = \pi\displaystyle \int_{-1}^{1} \left((3-y^2)^2-(3-(2-y^2))^2\right)dy=\pi\displaystyle \int_{-1}^{1} \left((9-6y^2+y^4)-(1+2y^2+y^4)\right) =\pi\displaystyle \int_{-1}^{1} \left(8-8y^2\right)dy=16\pi\displaystyle \int_{0}^{1} (1-y^2)dy=\dfrac{32\pi}{3}.$.

Answer 2

Not an answer but a reference for understanding.

$$ \text { Volume of revolution}=\int_{-1}^{1} \pi\left(3-2+y^{2}\right)^{2} d y-\int_{-1}^{1} \pi\left(3-y^{2}\right)^{2} d y $$