Volume integral $\int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos(t)+1}\>r dr dt$

Question

I need to calculate the volume of the solid $$E=\{(x,y,z)\in\mathbb{R}^{3}\mid x^2+y^2-2y\leq0,0\leq z\leq\sqrt{(x-1)^2+(y-1)^2}\}$$ I know $$x^2+y^2-2y=0$$ can be rewritten in polar coordinates $$x=r\cos(t),\quad y-1=r\sin(t)$$ with $0\leq r\leq1,0\leq t\leq2\pi$. Then the volume integral for the solid is $$\int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos(t)+1}\>r dr \space dt$$ But I have no idea how to evaluate this integral.

Answer 1

Observe that the resulting integral for the volume

$$V= \int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos t+1}\>r \space dr \space dt$$

is on the integrand $\sqrt{r^2-2r\cos t+1}$, which is the distance between the fixed point $(1,0)$ on the edge of the unit disk and any point $(r, \theta)$ over the unit disk. Then, with the polar origin chosen at the fixed point, the integral can be equivalently expressed as

$$V= \int_{-\frac\pi2}^{\frac\pi2}\int_{0}^{2\cos\theta}r\>rdr \space dt= \frac{16}3\int_{0}^{\frac\pi2}\cos^3tdt=\frac{32}9 $$

Answer 2

While I am late, wanted to share this answer. This is volume of cylinder $x^2+(y-1)^2 \leq 1$, above the plane $z = 0$ and below the surface of the cone $z = \sqrt{(x-1)^2+(y-1)^2}$.

As you can see, the vertex of the cone $(1, 1, 0)$ is on the surface of the cylinder. For integral, it is easier to parametrize the surface of the cone as $x = 1 + r\cos \theta, y = 1 + r\sin \theta, z = r, 0 \leq \theta \leq 2\pi$.

So the equation of the cylinder becomes, $r \leq -2 \cos \theta$ and forms between $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$.

$V = \displaystyle \int_{\pi/2}^{3\pi/2} \int_0^{-2\cos\theta} r^2 \ dr \ d\theta = \frac{32}{9}$