Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified linee

Question

• $y=x^2$, $x=y^2$ ; about $x=-1$.

My answer is: $$\pi \int _{0}^{1}\left( 1+\sqrt {y}\right) ^{2}-\left( 1+y\right) ^{2}dy.$$

Can you check my answer?

Answer 1

I find that I am in disagreement with your answer. In this problem we can say that

$$V=2\pi\int\int (x+1)dydx=2\pi\int_0^1\int_{x^2}^{\sqrt{x}}(x+1)dy\,dx=2\pi\int_0^1\int_{y^2}^{\sqrt{y}}(x+1)dx\,dy$$

As you can see, it can be integrated in either order; both give the same results. Following your lead, we'll integrate over $x$ first, thus

$$V=2\pi\int_0^1\left(\frac{x^2}{2}+x\right)\big|_{y^2}^{\sqrt{y}}\,dy=2\pi\int_0^1\left(\frac{y}{2}+\sqrt{y}-\frac{y^4}{2}-y\right)\,dy$$

Both of the integrals I showed have the same result. I have verified the volume calculation by an independent method, as well. I believe that the error you made was in saying the the integral of $(x+1)$ is equal to $\frac{(x+1)^2}{2}$.