1. The problem is to understand and compute surface and volume integrals. 2. Surface integrals calculate the integral of a function over a surface $S$. The formula is $$\iint_S f(x,y,z)\, dS$$ where $dS$ is the surface element. 3. Volume integrals calculate the integral of a function over a volume $V$. The formula is $$\iiint_V f(x,y,z)\, dV$$ where $dV$ is the volume element. 4. Important rules: - For surface integrals, parametrize the surface or use the vector form with the normal vector. - For volume integrals, use appropriate coordinate systems (Cartesian, cylindrical, spherical) depending on the volume shape. 5. Example: Compute the surface integral of $f(x,y,z) = x + y + z$ over the unit sphere $x^2 + y^2 + z^2 = 1$. 6. Parametrize the sphere using spherical coordinates: $$x = \sin\phi \cos\theta, y = \sin\phi \sin\theta, z = \cos\phi$$ with $\phi \in [0, \pi]$, $\theta \in [0, 2\pi]$. 7. The surface element on a sphere is $$dS = R^2 \sin\phi \, d\phi \, d\theta$$ with $R=1$. 8. Substitute $f$ and $dS$: $$\iint_S (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi \, d\phi \, d\theta$$ 9. Separate the integral: $$\int_0^{2\pi} \int_0^{\pi} (\sin\phi \cos\theta + \sin\phi \sin\theta + \cos\phi) \sin\phi \, d\phi \, d\theta$$ 10. Integrate with respect to $\theta$: - $\int_0^{2\pi} \cos\theta \, d\theta = 0$ - $\int_0^{2\pi} \sin\theta \, d\theta = 0$ - $\int_0^{2\pi} d\theta = 2\pi$ 11. The integral reduces to $$2\pi \int_0^{\pi} \cos\phi \sin\phi \, d\phi$$ 12. Use substitution $u = \sin^2\phi$, $du = 2\sin\phi \cos\phi \, d\phi$, so $$\int_0^{\pi} \cos\phi \sin\phi \, d\phi = \frac{1}{2} \int_0^{\pi} du = 0$$ because $\sin^2\phi$ is zero at both limits. 13. Therefore, the surface integral evaluates to 0. Final answer: The surface integral of $f(x,y,z) = x + y + z$ over the unit sphere is $0$.