Use cylindrical or spherical coordinates, whichever seems more appropriate. Find the volume V of the solid E that lies above the cone z=sqrt (x^2+y^2) and below the sphere x^2+y^2+z^2=1 v=square

1. Choose appropriate coordinates<br /> Spherical coordinates are suitable because the cone and sphere are symmetric about the origin. Use:<br />$x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$, and $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.<br /><br />2. Determine bounds for $\rho$, $\phi$, and $\theta$<br /> The cone $z = \sqrt{x^2 + y^2}$ implies $\tan\phi = 1$, so $\phi = \frac{\pi}{4}$. The sphere $x^2 + y^2 + z^2 = 1$ gives $\rho = 1$. Thus:<br />- $\rho$: $0 \leq \rho \leq 1$<br />- $\phi$: $\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}$<br />- $\theta$: $0 \leq \theta \leq 2\pi$<br /><br />3. Set up the integral<br /> The volume is given by:<br />$$ V = \int_{0}^{2\pi} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{1} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$<br /><br />4. Evaluate the integral over $\rho$<br /> Integrate with respect to $\rho$:<br />$$ \int_{0}^{1} \rho^2 \, d\rho = \left[\frac{\rho^3}{3}\right]_{0}^{1} = \frac{1}{3} $$<br /><br />5. Evaluate the integral over $\phi$<br /> Substitute and integrate:<br />$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin\phi \, d\phi = \left[-\cos\phi\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\cos\frac{\pi}{2} + \cos\frac{\pi}{4} = 0 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} $$<br /><br />6. Evaluate the integral over $\theta$<br /> Integrate over $\theta$:<br />$$ \int_{0}^{2\pi} 1 \, d\theta = 2\pi $$<br /><br />7. Combine results<br /> Multiply the results of all integrals:<br />$$ V = \frac{1}{3} \cdot \frac{\sqrt{2}}{2} \cdot 2\pi = \frac{\sqrt{2}\pi}{3} $$