# [SOLVED]Another surface integral

##### Member
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
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

$\chi$ $\sigma$

##### Member
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.

##### Member
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

##### Member
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
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.

##### Member
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