Integrate Hemisphere 7772B7

1. **State the problem:** We want to integrate the function $f(x,y,z) = z$ over a solid hemisphere using spherical coordinates. 2. **Recall spherical coordinates:** The spherical coordinates $(r, \theta, \phi)$ relate to Cartesian coordinates as: $$x = r \sin\phi \cos\theta, \quad y = r \sin\phi \sin\theta, \quad z = r \cos\phi$$ where $r \geq 0$ is the radius, $\theta$ is the azimuthal angle in $[0,2\pi]$, and $\phi$ is the polar angle from the positive $z$-axis in $[0,\pi]$. 3. **Describe the hemisphere:** The solid hemisphere above the $xy$-plane corresponds to $z \geq 0$, so $\phi$ ranges from $0$ to $\frac{\pi}{2}$. 4. **Set the limits:** For a hemisphere of radius $R$, the limits are: $$0 \leq r \leq R, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq \phi \leq \frac{\pi}{2}$$ 5. **Volume element in spherical coordinates:** The volume element is $$dV = r^2 \sin\phi \, dr \, d\phi \, d\theta$$ 6. **Rewrite the integral:** The integral of $f(x,y,z) = z = r \cos\phi$ over the hemisphere is $$\int_0^{2\pi} \int_0^{\pi/2} \int_0^R r \cos\phi \cdot r^2 \sin\phi \, dr \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\pi/2} \int_0^R r^3 \cos\phi \sin\phi \, dr \, d\phi \, d\theta$$ 7. **Integrate with respect to $r$:** $$\int_0^R r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4}$$ 8. **Integral becomes:** $$\frac{R^4}{4} \int_0^{2\pi} \int_0^{\pi/2} \cos\phi \sin\phi \, d\phi \, d\theta$$ 9. **Integrate with respect to $\phi$:** Use substitution $u = \sin\phi$, $du = \cos\phi d\phi$: $$\int_0^{\pi/2} \cos\phi \sin\phi \, d\phi = \int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$$ 10. **Integral becomes:** $$\frac{R^4}{4} \int_0^{2\pi} \frac{1}{2} \, d\theta = \frac{R^4}{8} \int_0^{2\pi} d\theta = \frac{R^4}{8} \cdot 2\pi = \frac{\pi R^4}{4}$$ **Final answer:** $$\boxed{\frac{\pi R^4}{4}}$$