91影视

To find the volume of the solid, we can use the following volume integral: $$ V = \int_{V} \rho^2 \sin{\varphi} d\rho d\varphi d\theta $$ Next, we must determine the limits of integration in terms of \(\rho\), \(\varphi\), and \(\theta\). For the radial coordinate \(\rho\), we are considering the solid inside the sphere \(\rho=1\), so the limits will be from 0 to 1. For the polar angle \(\varphi\), all points inside the solid are below the cone with \(\varphi=\pi/4\). Hence, \(\varphi\) will range from 0 to \(\pi/4\). For the azimuthal angle \(\theta\), the solid is restricted to non-negative \(z\)-values, which corresponds to the upper half-space. Therefore, \(\theta\) will range from 0 to \(2\pi\).

With the limits of integration set, we can compute the volume integral: $$ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^1 \rho^2 \sin\varphi \, d\rho d\varphi d\theta $$ Let's compute this step by step, starting with the radial integral: $$ \int_0^1 \rho^2 d\rho = \frac{1}{3}\rho^3 \bigg\rvert_0^1 = \frac{1}{3} $$ Next, let's compute the polar integral: $$ \int_0^{\pi/4} \sin\varphi \, d\varphi = -\cos\varphi \bigg\rvert_0^{\pi/4} = -\frac{\sqrt{2}}{2} + 1 $$ Finally, let's compute the azimuthal integral: $$ \int_0^{2\pi} \, d\theta = 2\pi $$ In order to find the volume of the solid, multiply these results together: $$ V = \frac{1}{3} \left(-\frac{\sqrt{2}}{2} + 1\right) (2\pi) = -\frac{\pi\sqrt{2}}{3} + \frac{2\pi}{3} $$ The volume of the solid inside the sphere \(\rho=1\) and below the cone \(\varphi=\pi / 4\) for \(z \geq 0\) is \(-\frac{\pi\sqrt{2}}{3} + \frac{2\pi}{3}\).