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[SOLVED] Another surface integral

MacLaddy

Member
Jan 29, 2012
52
Here is another that I am stuck on.

Please doublecheck my work, and let me know if where I am stuck is correct, or if I am on the completely wrong path.

Evaluate the surface integral \(\int\int f(x,y,z)dS\) using an explicit representation of the surface.

\(f(x,y,z) = x^2 + y^2;\mbox{ S is the paraboloid } z= x^2 + y^2\mbox{ for }0\leq z \leq 4\)

\(dS=\sqrt{4x^2+4y^2+1}\)

\(\int\int(x^2+y^2)*2*\sqrt{x^2+y^2+\frac{1}{4}}dA\)

\(\int_0^{2\pi}\int_0^4(r^2)*(r^2+\frac{1}{4})^{1/2}dzrdr\)

It's late, I'm not thinking straight. I'm sure that this should be integrated with respect to dxdy, not dzrdr, but it isn't clicking right.

A point in the right direction would be appreciated.

Thanks,
Mac
 

chisigma

Well-known member
Feb 13, 2012
1,704
Here is another that I am stuck on.

Please doublecheck my work, and let me know if where I am stuck is correct, or if I am on the completely wrong path.

Evaluate the surface integral \(\int\int f(x,y,z)dS\) using an explicit representation of the surface.

\(f(x,y,z) = x^2 + y^2;\mbox{ S is the paraboloid } z= x^2 + y^2\mbox{ for }0\leq z \leq 4\)

\(dS=\sqrt{4x^2+4y^2+1}\)

\(\int\int(x^2+y^2)*2*\sqrt{x^2+y^2+\frac{1}{4}}dA\)

\(\int_0^{2\pi}\int_0^4(r^2)*(r^2+\frac{1}{4})^{1/2}dzrdr\)

It's late, I'm not thinking straight. I'm sure that this should be integrated with respect to dxdy, not dzrdr, but it isn't clicking right.

A point in the right direction would be appreciated.

Thanks,
Mac
The general formula is...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = \int \int_{D} f \{x,y,g(x,y)\}\ \sqrt{1 + (\frac {\partial g}{\partial x})^{2} + (\frac {\partial g}{\partial y})^{2}}\ dx\ dy$ (1)

In Your case is $\displaystyle g(x,y)= x^{2} + y^{2}$ and D is the circle of radious 2 centered in x=y=0, so that is...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int \int_{D} (x^{2}+y^{2})\ \sqrt {\frac{1}{4} + x^{2} + y^{2}}\ dx\ dy$ (2)

Using polar coordinates $\rho$ and $\theta$ the integral becomes...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int_{0}^{2\ \pi} \int_{0}^{2} \rho^{3}\ \sqrt {\frac{1}{4} + \rho^{2}}\ d \theta\ d \rho = \frac{\pi}{60}\ |(4\ \rho^{2} + 1)^{\frac{3}{2}}\ (6\ \rho^{2} - 1)|_{0}^{2} $ (3)

Kind regards

$\chi$ $\sigma$
 

MacLaddy

Member
Jan 29, 2012
52
The general formula is...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = \int \int_{D} f \{x,y,g(x,y)\}\ \sqrt{1 + (\frac {\partial g}{\partial x})^{2} + (\frac {\partial g}{\partial y})^{2}}\ dx\ dy$ (1)

In Your case is $\displaystyle g(x,y)= x^{2} + y^{2}$ and D is the circle of radious 2 centered in x=y=0, so that is...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int \int_{D} (x^{2}+y^{2})\ \sqrt {\frac{1}{4} + x^{2} + y^{2}}\ dx\ dy$ (2)

Using polar coordinates $\rho$ and $\theta$ the integral becomes...

$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int_{0}^{2\ \pi} \int_{0}^{2} \rho^{3}\ \sqrt {\frac{1}{4} + \rho^{2}}\ d \theta\ d \rho = \frac{\pi}{60}\ |(4\ \rho^{2} + 1)^{\frac{3}{2}}\ (6\ \rho^{2} - 1)|_{0}^{2} $ (3)

Kind regards

$\chi$ $\sigma$
Ahh, good. So I was on the right track. It looks like I'm just using r where your using $\rho$.

Now I'll have to figure out how to evaluate that integral.

Thanks,
Mac

*EDIT* Your integral has different limits of integration. I'll have to dig into it and see if I can figure out why.
 

MacLaddy

Member
Jan 29, 2012
52
Actually, no. Now I'm doubly confused.

Why does your limit of integration $\rho$ have 0 to $2\pi$ as it's limits, and $d\theta$ have 0 to 2? Isn't $\rho$ just the radius of the paraboloid $0 \leq \rho \leq 2$? And $d\theta$ is the circle $0 \leq \theta \leq 2\pi$?

Doesn't the height, $0 \leq z \leq 4$ come into play at some point in the integration?

Thanks again,
Mac
 

MacLaddy

Member
Jan 29, 2012
52
Alright, not sure if anyone is still looking at this problem of mine, but I think I may have figured out the final solution. However, there is a bit of a trick I am doing here that I do not know if it's valid.


$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int \int_{D} (x^{2}+y^{2})\ \sqrt {\frac{1}{4} + x^{2} + y^{2}}\ dx\ dy$

$\displaystyle 2\ \int_0^{2\pi} \int_0^{2} (r^{2})\ \sqrt {\frac{1}{4} + r^2}\ rdr\ d\theta $

$\displaystyle 2\ \int_0^{2\pi} \int_0^{2} (r^{3})\ \sqrt {\frac{1}{4} + r^2}\ dr\ d\theta $

$\mbox{Let u =}\frac{1}{4}+r^{2}$

$du = 2rdr$

$ \frac{1}{2}du=rdr$

$ (\frac{1}{2})^3 du = r^{3}dr \mbox{ (This is the part that I don't know if it's valid)}$

$\displaystyle [2*\frac{1}{8}] \int_0^{2\pi} \int \sqrt {u}\ du\ d\theta $

This should simplify finally down to $\frac{4\pi}{3}$

If anyone could take a moment and let me know if that is correct I would be very appreciative.

Thank you,
Mac
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi MacLaddy. :)

Why does your limit of integration $\rho$ have 0 to $2\pi$ as it's limits, and $d\theta$ have 0 to 2? Isn't $\rho$ just the radius of the paraboloid $0 \leq \rho \leq 2$? And $d\theta$ is the circle $0 \leq \theta \leq 2\pi$?
I think there is a little typo. The integral in Chisigma's post #2 should be,

\[\int \int_{S} f(x,y,z)\ dS = 2\ \int_{0}^{2\ \pi} \int_{0}^{2} \rho^{3}\ \sqrt {\frac{1}{4} + \rho^{2}}\color{red}{d \rho\,d\theta} \]

Doesn't the height, $0 \leq z \leq 4$ come into play at some point in the integration?
On the paraboloidal surface the \(z\) coordinate can be given by \(x^2+y^2\) so if you have a \(z\) in your integration it can be replaced by \(x^2+y^2\).

Alright, not sure if anyone is still looking at this problem of mine, but I think I may have figured out the final solution. However, there is a bit of a trick I am doing here that I do not know if it's valid.


$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int \int_{D} (x^{2}+y^{2})\ \sqrt {\frac{1}{4} + x^{2} + y^{2}}\ dx\ dy$

$\displaystyle 2\ \int_0^{2\pi} \int_0^{2} (r^{2})\ \sqrt {\frac{1}{4} + r^2}\ rdr\ d\theta $

$\displaystyle 2\ \int_0^{2\pi} \int_0^{2} (r^{3})\ \sqrt {\frac{1}{4} + r^2}\ dr\ d\theta $

$\mbox{Let u =}\frac{1}{4}+r^{2}$

$du = 2rdr$

$ \frac{1}{2}du=rdr$

$ \color{red}{(\frac{1}{2})^3 du = r^{3}dr \mbox{ (This is the part that I don't know if it's valid)}}$

$\displaystyle [2*\frac{1}{8}] \int_0^{2\pi} \int \sqrt {u}\ du\ d\theta $

This should simplify finally down to $\frac{4\pi}{3}$

If anyone could take a moment and let me know if that is correct I would be very appreciative.

Thank you,
Mac
The highlighted part is incorrect. You cannot raise the power of parts of the equation. You can write, \( \frac{1}{2}du=rdr\Rightarrow \left( \frac{1}{2}\right)^3(du)^3=r^3 (dr)^3\) but I doubt whether this will be helpful in solving the integral.

Kind Regards,
Sudharaka.
 

MacLaddy

Member
Jan 29, 2012
52
Hi MacLaddy. :)



I think there is a little typo. The integral in Chisigma's post #2 should be,

\[\int \int_{S} f(x,y,z)\ dS = 2\ \int_{0}^{2\ \pi} \int_{0}^{2} \rho^{3}\ \sqrt {\frac{1}{4} + \rho^{2}}\color{red}{d \rho\,d\theta} \]



On the paraboloidal surface the \(z\) coordinate can be given by \(x^2+y^2\) so if you have a \(z\) in your integration it can be replaced by \(x^2+y^2\).



The highlighted part is incorrect. You cannot raise the power of parts of the equation. You can write, \( \frac{1}{2}du=rdr\Rightarrow \left( \frac{1}{2}\right)^3(du)^3=r^3 (dr)^3\) but I doubt whether this will be helpful in solving the integral.

Kind Regards,
Sudharaka.

Thanks, Sudharaka.

This is one of those problems that when finally worked through to the end, I end up kicking myself for making it far more difficult then it really was.

It's an off week, I think.

Thanks all,
Mac