Find the volume of the solid that lies under the paraboloid $z = 8x^2 + 8y^2$ above the $xy$-plane, and inside the cylinder $x^2 + y^2 = 2x$.
I am trying to figure out the double integral in terms of $r$ and I don't know why I am wrong.
This is what I wrote:
$$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}(8r^2)r dr d\theta.$$
Best Answer
You have everything:
$$V = \displaystyle\int_{-0.5\pi}^{0.5\pi}\int_{0}^{2\cos(\theta)}8r^3drd\theta$$ $$V = \displaystyle\int_{-0.5\pi}^{0.5\pi} 8\left( \frac{r^4}{4}\mid_0^{2\cos(\theta)}\right)d\theta$$ $$V = \displaystyle\int_{-0.5\pi}^{0.5\pi} 32 \cos^4(\theta)d\theta$$
$$V = 32\left( \frac{1}{32}(12\theta+8\sin(2\theta)+\sin(4\theta)) \right)\mid_{-0.5\pi}^{0.5\pi} = 12\pi $$