# Surface integral over an annulus

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• Aug 31st 2010, 04:55 AM
bobred
Surface integral over an annulus
Hi, would just like to know if what I have done is correct.
Find the area integral of the surface $z=y^2+2xy-x^2+2$ in polar form lying over the annulus $\frac{3}{8}\leq x^2+y^2\leq1$

The equation in polar form is $r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2$

$\displaystyle{\int^{\pi}_{-\pi}}\int^1_{\sqrt{\frac{3}{8}}}(r^2\sin^2\theta+2 r^2\cos\theta\sin\theta-r^2\cos^2\theta+2)\, r\, dr\, d\theta$

Integrating with respect to $r$

${\frac {55}{256}}\, \left( \sin \left( \theta \right) \right) ^{2}+{
\frac {55}{128}}\,\cos \left( \theta \right) \sin \left( \theta
\right) -{\frac {55}{256}}\, \left( \cos \left( \theta \right)
\right) ^{2}+5/8
$

and with respect to $\theta$, I get the area as

$\frac{5}{4}\pi$

Does this look ok?
• Aug 31st 2010, 08:52 AM
Opalg
Quote:

Originally Posted by bobred
Does this look ok?

Yes. (Nod)
• Aug 31st 2010, 08:58 AM
bobred
Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.

James
• Aug 31st 2010, 09:56 AM
AllanCuz
Quote:

Originally Posted by bobred
Thanks for checking it over I had some post that said there was a problem, that what I was working out was a volume.

James

That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.
• Aug 31st 2010, 10:52 AM
wonderboy1953
Diamond transform
Quote:

Originally Posted by AllanCuz
That person probably thinks that any triple integral expresses volume and only volume. It's hard for some people to understand that a single, double or triple integral can express both area and volume. After all, a double integral is a simplified triple integral and a single integral is a simplified double integral.

Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).
• Aug 31st 2010, 12:59 PM
Opalg
Quote:

Originally Posted by bobred
Find the area integral of the surface $z=y^2+2xy-x^2+2$ in polar form lying over the annulus $\frac{3}{8}\leq x^2+y^2\leq1$

Quote:

Originally Posted by bobred
I had some post that said there was a problem, that what I was working out was a volume.

Oops, I think there is indeed a problem. (Doh)

What you did was to find the volume between the annulus and the surface. But it looks as though the question was actually asking for the surface area of that part of the suface lying above the annulus.

The formula for surface area says that this area is given by $\displaystyle\iint_A\sqrt{\bigl(\tfrac{\partial z}{\partial x}\bigr)^2 + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2+1}\,dxdy$, where A denotes the annulus.

If $z = y^2 + 2xy - x^2+2$ then $\tfrac{\partial z}{\partial x} = 2(y-x)$, $\tfrac{\partial z}{\partial y} = 2(y+x)$, and $\bigl(\tfrac{\partial z}{\partial x}\bigr)^2} + \bigl(\tfrac{\partial z}{\partial y}\bigr)^2} = 8(x^2+y^2) = 8r^2$. So the answer to the question is

. . . . . \begin{aligned}\iint_A\sqrt{8r^2+1}\,rdrd\theta &= \int_0^{2\pi}\int_{\sqrt{3/8}}^1\sqrt{8r^2+1}\,rdrd\theta \\ &= \int_0^{2\pi}\Bigl[\tfrac1{24}(8r^2+1)^{3/2}\Bigr]_{\sqrt{3/8}}^1\,d\theta \\ &= \int_0^{2\pi}\!\!\tfrac{19}{24}\,d\theta = \tfrac{19}{12}\pi.\end{aligned}

I thought that the question was not very clearly worded, but the fact that those 3/2 powers both turn out to be integers makes me fairly sure that "area integral" is supposed to mean "surface area" in this question.
• Aug 31st 2010, 01:09 PM
bobred
Thanks, I had been looking for an expression of the form $(x^2+y^2)=r^2$ because the trig expression was horrendous.

James
• Aug 31st 2010, 06:50 PM
AllanCuz
Quote:

Originally Posted by wonderboy1953
Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).

You've been out of college for 25 years and you're alias is wonderBOY?!
• Sep 1st 2010, 12:31 AM
Opalg
Quote:

Originally Posted by AllanCuz
Quote:

Originally Posted by wonderboy1953
Since it's been over 25 years I've been in college, I'd like to see the formula that converts a double integral into a single integral (which I understand is called a diamond transform).

You've been out of college for 25 years and you're alias is wonderBOY?!

That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday. (Rofl)
• Sep 1st 2010, 06:14 PM
AllanCuz
Quote:

Originally Posted by Opalg
That didn't come as a surprise to me. Wonderboy1953 clearly wasn't born yesterday. (Rofl)

I was never one for numbers!
• Sep 2nd 2010, 07:40 AM
wonderboy1953
Quote:

Originally Posted by AllanCuz
I was never one for numbers!

Since you were never one for numbers, what brings you to a math website?

Since this is the second time (besides this thread) that I haven't gotten an answer to my question leads me to believe that no such formula exists to transform a double integral into a single integral. Can you prove me wrong?
• Sep 2nd 2010, 08:03 AM
Jester
In some cases doesn't Green's theorem do that?
• Sep 2nd 2010, 08:08 AM
wonderboy1953
Quote:

Originally Posted by Danny
In some cases doesn't Green's theorem do that?

Doesn't that deal with line integrals?
• Sep 2nd 2010, 08:11 AM
Ackbeet
Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.
• Sep 2nd 2010, 08:15 AM
wonderboy1953
Quote:

Originally Posted by Ackbeet
Green's Theorem relates a surface integral to a line integral. Since surface integrals often decompose into a double iterated integral, and line integrals often decompose into a single integral, you could say that Green's Theorem can reduce some double integrals into a single integral. I'm not sure there is a way, short of actually computing the inner integral, to reduce all double integrals to single integrals.