25. find the volume of the solid that is enclosed by the cone z = √(x² + y²) and the sphere x² + y² + z² = 2.

Explanation:

Step1: Convert to spherical coordinates

In spherical coordinates, $x = \rho\sin\varphi\cos\theta$, $y=\rho\sin\varphi\sin\theta$, $z = \rho\cos\varphi$. The cone $z=\sqrt{x^{2}+y^{2}}$ gives $\rho\cos\varphi=\rho\sin\varphi$, so $\tan\varphi = 1$ and $\varphi=\frac{\pi}{4}$. The sphere $x^{2}+y^{2}+z^{2}=2$ gives $\rho^{2}=2$, so $\rho=\sqrt{2}$. The volume element in spherical - coordinates is $dV=\rho^{2}\sin\varphi d\rho d\varphi d\theta$.

Step2: Set up the triple - integral for volume

The limits of integration are: $0\leqslant\theta\leqslant2\pi$, $0\leqslant\varphi\leqslant\frac{\pi}{4}$, $0\leqslant\rho\leqslant\sqrt{2}$. The volume $V=\int_{\theta = 0}^{2\pi}\int_{\varphi = 0}^{\frac{\pi}{4}}\int_{\rho = 0}^{\sqrt{2}}\rho^{2}\sin\varphi d\rho d\varphi d\theta$.

Step3: Integrate with respect to $\rho$

First, integrate $\int_{\rho = 0}^{\sqrt{2}}\rho^{2}d\rho=\left[\frac{1}{3}\rho^{3}\right]{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{3}$. The integral becomes $V=\frac{2\sqrt{2}}{3}\int{\theta = 0}^{2\pi}\int_{\varphi = 0}^{\frac{\pi}{4}}\sin\varphi d\varphi d\theta$.

Step4: Integrate with respect to $\varphi$

Integrate $\int_{\varphi = 0}^{\frac{\pi}{4}}\sin\varphi d\varphi=[-\cos\varphi]{0}^{\frac{\pi}{4}}=1 - \frac{\sqrt{2}}{2}$. The integral becomes $V=\frac{2\sqrt{2}}{3}(1 - \frac{\sqrt{2}}{2})\int{\theta = 0}^{2\pi}d\theta$.

Step5: Integrate with respect to $\theta$

Integrate $\int_{\theta = 0}^{2\pi}d\theta=[\theta]_{0}^{2\pi}=2\pi$. Then $V=\frac{2\sqrt{2}}{3}(1 - \frac{\sqrt{2}}{2})\times2\pi=\frac{4\pi}{3}(\sqrt{2}-1)$.

Answer:

$\frac{4\pi}{3}(\sqrt{2}-1)$