Stoke’s Theorem

Solved Problems

Solution.

We suppose that $$ is the part of the plane cut by the cylinder. The curve $$ is oriented counterclockwise when viewed from the end of the normal vector $$ which has coordinates

As $$ $$ $$ we can write:

Applying Stoke's Theorem, we find:

We can express the surface integral in terms of the double integral:

The equation of the plane is $$ so the square root in the integrand is equal to

Hence,

The region $$ is the circle $$ of radius $$ By changing to polar coordinates, we get

Solution.

Let the surface $$ be the part of the plane $$ bounded by the ellipse. Obviously that the unit normal vector is $$ Since

then the curl of the vector field $$ is

By Stoke's Theorem,

The double integral in the latter formula is the area of the ellipse. Therefore, the integral is

Solution.

We suppose that the surface $$ is the plane of the triangle $$ Orientation of the surface $$ and the contour $$ are shown in Figure $$

We first find the unit normal vector $$

Then

and hence,

In our case $$ $$ $$ so the curl of $$ is

By Stoke's formula,

Here the double integral $$ is the area of the triangle $$ which is equal to

The complete answer is

Solution.

Let $$ be the part of the plane cut by the paraboloid. Orientation of the surface $$ and the curve $$ are shown in Figure $$

The normal vector $$ can be found from the equation of the plane:

Since

the curl of the vector field $$ is

By Stoke's formula, we have

As $$ the integral becomes

To complete the calculation, we must evaluate the double integral $$ i.e. the area of the surface $$ The explicit equation of the plane is $$ Therefore, using the formula

where $$ is projection of $$ onto the $$-plane, we have

Determine the region of integration $$ Solving the system of the equations

we obtain

Thus, we see that the region $$ is the circle of radius $$ centered at $$ Then the area of the region $$ is

Hence, the initial integral is