Surface Integral of Two Surfaces

In summary, to solve this question, you need to parametrize the cone and use the conditions given by the cylinder to determine the limits of integration for u and v, which will then allow you to evaluate the surface integral using the parametrization.
  • #1
Lomion
9
0
Hello!

This is a question from one of our past exams, and it's had me stumped for the past hour. The question states:

The cylinder [tex]x^2+y^2=2x[/tex] cuts out a portion of a surface S from the upper nappe of the cone [tex]x^2+y^2=z^2[/tex].

Compute the surface integral: [tex]\int\int (x^4-y^4+y^2z^2-z^2x^2+1) dS[/tex]

I'm mainly having trouble getting started. What exactly is the surface that we're supposed to evaluate the integral over?

My guess on this question is that I should parametrize the cone:
[tex]T(u,v) = (vcosu, vsinu, v)[/tex]

And use that to find [tex]T_u X T_v[/tex].

But what do I do after this? In order to find the limits of integration for u and v, do I use the conditions given by the cylinder? [tex]v = 2cosu[/tex]?

Using that still doesn't give me the numerica limits for v, though.

Help, anyone?
 
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  • #2
To answer your question, yes, you should use the conditions given by the cylinder to get the limits of integration for u and v. The equation x^2+y^2=2x can be rewritten as 2cos^2u + 2sin^2u = 2vcosu. Since v is a function of u, the limits for u can be determined by solving this equation for u. That is, the limits of u will be the domain of the equation 2cos^2u + 2sin^2u = 2vcosu. From there, the limits of v can be determined by substituting the limits of u back into the equation x^2+y^2=2x. Once you have the limits of u and v, you can then parametrize the cone with T(u,v) = (vcosu, vsinu, v), calculate T_u X T_v and use that to evaluate the surface integral.
 
  • #3


Hi there!

The surface that we are supposed to evaluate the integral over is the surface S that is cut out by the cylinder x^2+y^2=2x from the upper nappe of the cone x^2+y^2=z^2. In order to evaluate the integral, we first need to find the parametrization of S.

As you have correctly guessed, we can use the parametrization of the cone T(u,v) = (vcosu, vsinu, v) to find the parametrization of S. To do this, we need to find the intersection curve of the cone and the cylinder. This curve can be parametrized as C(t) = (cos t, sin t, 2cos t).

Next, we can use the parametrization of C(t) to find the parametrization of S. This can be done by setting z = 2cos t in the parametrization of the cone T(u,v). This gives us T(u,v) = (vcosu, vsinu, 2cosu).

Now, we can find the surface integral by using the formula:

\int\int f(x,y,z) dS = \int\int f(T(u,v)) ||T_u X T_v|| dA

where T_u X T_v is the cross product of the partial derivatives of T with respect to u and v, and dA is the area element in the u-v plane.

To find the limits of integration for u and v, we can use the conditions given by the cylinder, x^2+y^2=2x. Since we have already parametrized the curve of intersection as C(t) = (cos t, sin t, 2cos t), we can use the limits of t to find the limits for u and v. In this case, t ranges from 0 to 2pi, so u ranges from 0 to 2pi and v ranges from 0 to 2.

I hope this helps! Let me know if you have any further questions. Good luck on your exam!
 

What is a surface integral of two surfaces?

A surface integral of two surfaces is a mathematical calculation that determines the total flux or flow of a vector field through the region enclosed by two surfaces in three-dimensional space. It takes into account the direction and magnitude of the vector field and the orientation of the surfaces.

What are the applications of surface integral of two surfaces?

The surface integral of two surfaces has various applications in physics, engineering, and mathematics. It is used to calculate the total work done by a force, the total mass of a three-dimensional object, and the total amount of fluid flowing through a region. It is also used in electromagnetic theory to calculate the total electric or magnetic flux through a surface.

How is a surface integral of two surfaces calculated?

The surface integral of two surfaces is calculated by dividing the region enclosed by the two surfaces into small sections, calculating the flux through each section, and then adding up all the individual fluxes. This process is known as integration and requires advanced mathematical techniques such as double integration.

What is the difference between a surface integral of two surfaces and a line integral?

A surface integral of two surfaces calculates the flux through a three-dimensional region, while a line integral calculates the work done along a one-dimensional path. Additionally, a surface integral takes into account the orientation and magnitude of the vector field, while a line integral only considers the direction of the vector field.

What are some common examples of surface integrals of two surfaces?

One common example of a surface integral of two surfaces is calculating the total mass of a three-dimensional object, such as a cone or sphere. Another example is calculating the total amount of fluid flowing through a pipe with varying cross-sectional areas. Surface integrals are also commonly used in electromagnetics to calculate the total electric or magnetic flux through a surface.

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