Surface Integral of Two Surfaces

[Lomion]

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 $$x^2+y^2=2x$$ cuts out a portion of a surface S from the upper nappe of the cone $$x^2+y^2=z^2$$.

Compute the surface integral: $$\int\int (x^4-y^4+y^2z^2-z^2x^2+1) dS$$

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:
$$T(u,v) = (vcosu, vsinu, v)$$

And use that to find $$T_u X T_v$$.

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? $$v = 2cosu$$?

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

Help, anyone?

 

[member 553083]

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.

 

[wais]

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!