Tutorial 4: Surface Integral
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- Evaluate the surface integral of the vector field over the surface that is the graph of over the rectangle .
Solution
Use the formula for a surface integral over a graph :
For β , therefore,
- Let be the triangle with vertices , and and let , calculate the surface integral
If the triangle is oriented by the βdownwardβ normal.
Solution
Since lies in a plane, it is part of the graph of a linear function .
Substituting the vertices of the triangle for , we get the equation
which we can solve to find , , , i.e., . We may take and as parameters,
or . The domain of the parametrization is the triangle with vertices at , , and in the plane. For this parametrization,
Since the third component of this vector is positive, the orientation determined by is βupwardβ, so we will have to multiply our find answer by to get the surface integral with the downward orientation.
Now, we have (with the minus sign reminding us that the orientation is wrong),
To compute the double integral, we draw the integtation domain in the -plane, in the left hand part of the Figure. By reduction to iterated integrals,
Carrying out the -integration, we get
- The equations , describe a disk of radius 5 lying in the plane . Suppose that is the position vector field . Compute .
Solution
Since the disk is parallel to the plane, the outward unit normal is . Hence and so . Thus,
Alternatively we may solve this problem by using the formula for surface integrals
over graphs:
With and the disk , we get
- Let be the closed surface that consists of the hemisphere , and its base , . Let be the electric field defined by . Find the electric flux across .
Solution
Write where is the upper hemisphere and is the disk. Hence
(i) Let be the unit normal pointing outward from . Then
(ii) The unit normal is and on . Hence,
Therefore,
- Find the area of the ellipse cut on the plane by the circular cylinder .
Solution
The surface lies in the plane so we use this to calculate . Differentiating the equation for the plane with respect to gives,
Differentiating the equation for the plane with respect to gives,
Hence,
Then the area of is found be calculating the surface integral over for the function . The projection of the surface, , onto the xy-plane is given by . Hence the area of is given by
Note, since is a circle or radius 1 centred at the area of is the area of a unit circle which is .
- Find the integral , where the surface is the part of the sphere lying in the first octant.
Solution
It is convenient to solve this integral in spherical coordinates. The area element for spherical surface is . As , we can write the integral in the following form
The domain of integration is defined as
Hence, the integral is
- Find the integral where is the part of the cylindrical surface parameterized by , , .
Solution
Calculate the partial derivatives,
and their cross product,
Then the area element of the given surface is
Now we can calculate the surface integral: