Surface integral that has me stumped.
Q: Calculate $\int \int_{S} F \cdot dA$
Where $F(x,y,z)= xi+yj+zk$ S is the boundary of the region $x^{2}+y^{2} \leq z \leq (2-x^{2}-y^{2})^{1/2}$ oriented so that the normal points out of the region.
Thx this is a good deal harder then the other recommended problems i have completed thx for the help.
This looks like a standard "applying the divergence theorem" problem. Notice $$ \iint_S \mathbf{F} \cdot d\mathbf{A} = \iiint_V \nabla\cdot \mathbf{F}\, dV = 3\iiint_V\,dV $$ The region $$ x^{2}+y^{2} \leq z \leq (2-x^{2}-y^{2})^{1/2} $$ which is the intersection of the paraboloid and a sphere looking like the following figure:
Notice in this figure the region is artificially divided into two parts using a cone.
First part is below the sphere $x^2+y^2+z^2 = 2$, above the cone $z = \sqrt{x^2+y^2}$, integral within the first part is easier to be integrated in spherical coordinates: $$ \begin{aligned} &r = \sqrt{x^2+y^2+z^2} \in (0,\sqrt{2}) \\ &\phi = \arccos\frac{z}{r} \in (0,\frac{\pi}{4}) \\ &\theta = \arctan\frac{y}{x} \in (0,2\pi) \end{aligned} $$ this is a cone with a spherical top. The other part is below the cone $z = \sqrt{x^2+y^2}$, above the paraboloid $z = x^2+y^2$. Integral within the second part is easier to be integrated in cylindrical coordinates: $$ \begin{aligned} &\rho = \sqrt{x^2+y^2} \in (0,1) \\ &z \in (\rho^2,\rho) \\ &\theta = \arctan\frac{y}{x} \in (0,2\pi) \end{aligned} $$ The integral becomes: $$ \iint_S \mathbf{F} \cdot d\mathbf{A} = 3\int^{2\pi}_0\int^{\frac{\pi}{4}}_0 \int_0^\sqrt{2} r^2 \,dr d\phi d\theta + 3\int^{2\pi}_0\int^{1}_0 \int_{\rho^2}^\rho \rho \,dz d\rho d\theta= \sqrt{2}\pi^2 + \frac{\pi}{2} $$