Surface Integral Homework: Evaluate ∫∫σ

Baumer8993
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Homework Statement


Evaluate ∫∫σ where S is a surface with sides S1, S2, and S3. S1 is a portion of the cylinder x2+y2 = 1 whose bottom S2 is the disk x2+y2 ≤ 1 and whose top S3 is the portion of the plane z = 1 + x that lies above S2.


Homework Equations


Surface integrals, and vector calculus.


The Attempt at a Solution


I am more stuck with starting this problem. Do I need to do two surface integrals, or something else? I am just completely lost on what to do here...
 
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Baumer8993 said:

Homework Statement


Evaluate ∫∫σ where S is a surface with sides S1, S2, and S3. S1 is a portion of the cylinder x2+y2 = 1 whose bottom S2 is the disk x2+y2 ≤ 1 and whose top S3 is the portion of the plane z = 1 + x that lies above S2.


Homework Equations


Surface integrals, and vector calculus.


The Attempt at a Solution


I am more stuck with starting this problem. Do I need to do two surface integrals, or something else? I am just completely lost on what to do here...

What are you integrating over the surface? You can do it by integrating over the three surfaces and adding them or you might replace the surface integral with a volume integral using the divergence theorem if you just need the total over all three surfaces. Just start doing something.
 
I though I could only use the divergence theorem for flux?
 
Baumer8993 said:
I though I could only use the divergence theorem for flux?

Of course, if it's not a flux integral then you can't use the divergence theorem. That's why I was asking WHAT you are integrating.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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