Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$

Question

I set up the triple integral as follows:

$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\sqrt{1-r^2}r=r^2)drd\theta$$$$2\pi(\frac{-1}{2})[\frac{2}{3}(1-r^2)^{3/2}-1]|^1_0=-\pi[(0-1)-(\frac{2}{3}-1)]=\pi+\frac{1}{3}$$

The answer should be $\frac{\pi}{3}(2-\sqrt{2})$

Answer 1

The region $E$ in cylindrical coordinate is given by $$ E = \left\{(r,\theta,z)\colon 0\le r\le 1/\sqrt{2}, 0\le\theta\le 2\pi, r\le z\le \sqrt{1-r^2}\right\}. $$ To see this, observe that $z$ is bounded below by the given cone means that $$ z\ge \sqrt{x^2+y^2} = r,$$ and $z$ is bounded above by the sphere of radius 1 means that $$ z\le\sqrt{1-x^2-y^2} = \sqrt{1-r^2}.$$ Moreover, the cone intersects the sphere when $x^2+y^2=z^2=1-x^2+y^2$, i.e. $$r^2=1-r^2\implies 2r^2=1\implies r=\pm\frac{1}{\sqrt{2}}.$$

Thus, the volume of solid over $E$ is given by \begin{align*} V = \int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\int_r^{\sqrt{1-r^2}} rdzdrd\theta & = 2\pi \int_0^{\frac{1}{\sqrt{2}}}\int_r^{\sqrt{1-r^2}} rdzdr\\ & = 2\pi\int_0^{\frac{1}{\sqrt{2}}} \left(r\sqrt{1-r^2} - r^2\right)\, dr. \end{align*} The second integral is $$ \int_0^{\frac{1}{\sqrt{2}}} r^2\, dr = \frac{r^3}{3}\bigg|_0^{\frac{1}{\sqrt{2}}} = \frac{1}{6\sqrt{2}}$$ while the first integral is (upon integration by substitution) \begin{align*} \int_0^{\frac{1}{\sqrt{2}}} r\sqrt{1-r^2}\, dr & = -\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)(1-r^2)^{3/2}\bigg|_0^{\frac{1}{\sqrt{2}}}\\ & = -\frac{1}{3}\left(\frac{1}{2\sqrt{2}}-1\right)\\ & = \frac{2\sqrt{2}-1}{6\sqrt{2}}. \end{align*} Hence, \begin{align*} V=2\pi\left(\frac{2\sqrt{2}-1}{6\sqrt{2}}-\frac{1}{6\sqrt{2}}\right) & = 2\pi\left(\frac{2\sqrt{2}-2}{6\sqrt{2}}\right) \\ & = 2\pi\left(\frac{\sqrt{2}-1}{3\sqrt{2}}\right) \\ & = \frac{2\pi}{3}\left(1-\frac{1}{\sqrt{2}}\right) \\ & = \frac{\pi}{3}(2-\sqrt{2}). \end{align*}

Answer 2

Note that the solid is of revolution, formed by rotating the circular sector between $z=\sqrt {1-x^2}$ and $z=x$ (about the $z$-axis).
From cylinder method, we have that \begin{align}V&=2\pi \int_{0}^{\sqrt2 \over 2}x|\sqrt {1-x^2}-x|dx\\ &=2\pi({1\over3} - {\sqrt2\over6}) \end{align}