I have the following two "calculate the volume of the solid obtained by rotating the region" questions.
Let $y = 2 – x^2$ and $y = x^4 – x^2$, and let $R$ denote the region bounded by the two curves. Find the volume of the solid obtained by rotating the region $R$ about the $y$-axis.
Find the volume of the solid obtained by rotating the unit circle $x^2 + y^2 = 1$ about the line $y = -2$.
For the second one, I have the following integral:
$$\pi \int^1_{-1} (\sqrt{1-x^2} – 2)^2 – (-2 – \sqrt{1-x^2})^2 dx $$
which ultimately evaluates to $4\pi^2$. Is this correct?
However, I am having difficulties setting up the integral for the first one. Since I need to rotate about the $y$-axis, I think I need to solve the equations in terms of $x$, but I'm not seeing how to do that with the second equation. I think my limits of integration should be $0$ and $2^{1/4}$. How can I go about setting up this integral?
Best Answer
For the second one, it should be
$ \displaystyle \pi \int^1_{-1} ((\sqrt{1-x^2} + 2)^2 - (2 - \sqrt{1-x^2})^2)~ dx$
What you have written will give $ - 4 \pi^2$.
For the first one, it is easier to apply cylindrical shell method.
At the intersection of both curves,
$y = x^4 - x^2 = 2 - x^2 \implies x = 2^{1/4}~$, to the right of y-axis.
So the integral is,
$ \displaystyle \int_0^{2^{1/4}} ((2 - x^2) - (x^4 - x^2)) \cdot 2 \pi x ~ dx = \frac{4 \sqrt2~ \pi}{3}$