Find the center of mass of the hemisphere $x^2+y^2+z^2=a^2, z \geqslant 0$, if it has constant density.

Evaluate the surface integral $\iint_S \mathbf{F} \cdot d \mathbf{S}$ for the given vector field $\mathbf{F}$ and the oriented surface $S$. In other words, find the flux of $\mathbf{F}$ across $S$. For closed surfaces, use the positive (outward) orientation.

$\mathbf{F}(x, y, z)=y \mathbf{j}-z \mathbf{k}$, $S$ consists of the paraboloid $y=x^2+z^2, 0 \leqslant y \leqslant 1$, and the disk $x^2+z^2 \leqslant 1, y=1$

Verify Stokes’ theorem. Assume that the surface S is oriented upward. $\mathbf{F}=2 z \mathbf{i}-3 x \mathbf{j}+4 y \mathbf{k} ; S$ that portion of the paraboloid $z=16-x^{2}-y^{2}$ for $z \geq 0$.

Use the divergence theorem to compute flux integral $\iint_{S} \mathbf{F} \cdot d \mathbf{S}$ where $\mathbf{F}(x, y, z)=y \mathbf{j}-z \mathbf{k}$ and S consists of the union of paraboloid $y=x^{2}+z^{2}, 0 \leq y \leq 1$, and disk $x^{2}+z^{2} \leq 1, y=1$, oriented outward. What is the flux through just the paraboloid?

F(x,y,z)=yzi+xzJ+zxk, S is the part of the paraboloid z=4-x^2-y^2 that lies above the square 0<=x<=1, 0<=y<=1, and has upward orientation

Use Stokes’ Theorem to evaluate ∫c F · dr, where F(x, y, z) = xyi + yzj + zxk, and C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), oriented counter clockwise as viewed from above.

Find the area of the surface.
The part of the paraboloid $z=4-x^2-y^2$ that lies above the x y-plane

Evaluate the surface integral

$$\iint_S y d S$$

where $S$ is the surface $z=\frac{2}{3}\left(x^{3 / 2}+y^{3 / 2}\right), 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1$.

Find the area of the surface.

The part of the plane 2x+3y-z=1 that lies above the rectangular region [1,2] x [1,3]

Evaluate $\iint_Sf(x,y,z)\ dS$.

f(x,y,z)= xy; S is the part of the plane 2x + 3y +z =6 in the first octant

The surface with parametric equations x=u^2, y=uv, z=1/2v^2, 0<=u<=1, 0<=v<=1

Find the area of the surface. The part of the surface z=xy that lies within the cylinder x^2+y^2=1

Find the area of the part of the plane $x + 2y + 3z = 1$ that lies inside the cylinder $x^2 + y^2 = 3$.

In given problem, find the area of the indicated surface. Make a sketch in each case.

The part of the plane 3x-2y+6z=12 that is bounded by the planes x=0, y=0, and 3x+2y=12