# [SOLVED]Surface Integral

##### Member
Hello all,

I've hit a roadblock on a question regarding Surface Integrals. I seem to be having a problem conceptualizing many of these concepts. Anyway, here goes.

Find the area of the following surface using an explicit description of a surface.

The cone $$z^2 = 4x^2 + 4y^2$$ for $$0\leq z\leq4$$

I have solved for z and found the dS portion to = $$\sqrt{5}$$ dA. This jives with the answer sheet that I have. It's the rest that I seem to be having troubles with.

Since it's a cone, shouldn't I just integrate like so?

$$\sqrt{5}\int_0^{2\pi}\int_0^4 (1)dz rdr \mbox{ for a final answer of }8\pi^2\sqrt5 \mbox{? The solution manual is showing an answer of }4\sqrt5$$.

Any assistance, or a point in the right direction, will be greatly appreciated.

Thanks,
Mac

#### CaptainBlack

##### Well-known member
Hello all,

I've hit a roadblock on a question regarding Surface Integrals. I seem to be having a problem conceptualizing many of these concepts. Anyway, here goes.

Find the area of the following surface using an explicit description of a surface.

The cone $$z^2 = 4x^2 + 4y^2$$ for $$0\leq z\leq4$$

I have solved for z and found the dS portion to = $$\sqrt{5}$$ dA. This jives with the answer sheet that I have. It's the rest that I seem to be having troubles with.

Since it's a cone, shouldn't I just integrate like so?

$$\sqrt{5}\int_0^{2\pi}\int_0^4 (1)dz rdr \mbox{ for a final answer of }8\pi^2\sqrt5 \mbox{? The solution manual is showing an answer of }4\sqrt5$$.

Any assistance, or a point in the right direction, will be greatly appreciated.

Thanks,
Mac
First observe that you have a cone with slant height $$2 \sqrt{5}$$ and base radius $$2$$, so the curved surface area is the circumference of the base ($$4 \pi$$) times the slant heigh over two = $$4 \pi \sqrt{5}$$.

Now lets do this using integration. First switch to cylindrical polars, then the surface element is $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$, so the area is:

$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} \frac{z}{2} d\theta dz$

etc...

(you should always draw a picture for these problems if you can)

Last edited:

##### Member
First observe that you have a cone with slant height $$2 \sqrt{5}$$ and base radius $$2$$, so the curved surface area is the circumference of the base ($$4 \pi$$) times the slant heigh over two = $$4 \pi \sqrt{5}$$.

Now lets do this using integration. First switch to cylindrical polars, then the surface element is $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$, so the area is:

$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz$

etc...

(you should always draw a picture for these problems if you can)
Thanks Captain, I appreciate the assistance.

I do have a couple of follow up questions, as I'm still not quite grasping this as well as I should.

I never actually took geometry, which is funny considering I'm now in Multivariate Calculus, but I haven't really needed it up to this point. So even though I know and can figure things out with simple geometry, it isn't usually the first thing that pops in my head when these problems arise.

I am still having some problems understanding this line,

Now lets do this using integration. First switch to cylindrical polars, then the surface element is $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$, so the area is:

$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz$
When doing cylindrical integrals in polar, I was under the impression that I had to use a triple integral? Also, for this particular question above my first instinct would be to use spherical integration, but I guess that would just find the area of the cone?

Can you clarify where you found $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$? Also, why is there an extra factor of "z" in your integrand? Is that similar to the Jacobian rdr, but zdz instead?

To assist you in assisting me, the things I do understand are, of course, the radius, the slant height (or hypotenuse in my non-geometry world). I also see that $$z^2 = 4x^2 + 4y^2\mbox{ can be worked to equal }\frac{z}{2}=r$$, but I don't know if that is where you came up with $$(z/2)\;d\theta$$, or how it applies.

Thanks again, I appreciate the assistance.

Mac

#### CaptainBlack

##### Well-known member
Thanks Captain, I appreciate the assistance.

I do have a couple of follow up questions, as I'm still not quite grasping this as well as I should.

I never actually took geometry, which is funny considering I'm now in Multivariate Calculus, but I haven't really needed it up to this point. So even though I know and can figure things out with simple geometry, it isn't usually the first thing that pops in my head when these problems arise.

I am still having some problems understanding this line,
You draw a picture of the surface element at a point on the surface where angle subtended by the element is $$\Delta \theta$$ and height is $$\Delta z$$ and calculate its area for small $$\Delta$$s (so they are replaced by $$d\theta$$ and $$dz$$ )

When doing cylindrical integrals in polar, I was under the impression that I had to use a triple integral? Also, for this particular question above my first instinct would be to use spherical integration, but I guess that would just find the area of the cone?

Can you clarify where you found $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$? Also, why is there an extra factor of "z" in your integrand? Is that similar to the Jacobian rdr, but zdz instead?
Yes, as you would be able to see if you draw a picture.

To assist you in assisting me, the things I do understand are, of course, the radius, the slant height (or hypotenuse in my non-geometry world). I also see that $$z^2 = 4x^2 + 4y^2\mbox{ can be worked to equal }\frac{z}{2}=r$$, but I don't know if that is where you came up with $$(z/2)\;d\theta$$, or how it applies.

Thanks again, I appreciate the assistance.
The area of a surface patch is $$r\;d\theta dl$$ where $$dl$$ is the distance in the surface between $$z$$ and $$z+dz$$ and you replace $$r$$ by its equivalent in terms of $$z$$.

As I have said repeatedly draw a picture showing the area element on the surface between $$\theta$$ and $$\theta+\Delta \theta$$ and $$z$$ and $$z+\Delta z$$ and claculate its area.

CB

##### Member
You draw a picture of the surface element at a point on the surface where angle subtended by the element is $$\Delta \theta$$ and height is $$\Delta z$$ and calculate its area for small $$\Delta$$s (so they are replaced by $$d\theta$$ and $$dz$$ )

Yes, as you would be able to see if you draw a picture.

The area of a surface patch is $$r\;d\theta dl$$ where $$dl$$ is the distance in the surface between $$z$$ and $$z+dz$$ and you replace $$r$$ by its equivalent in terms of $$z$$.

As I have said repeatedly draw a picture showing the area element on the surface between $$\theta$$ and $$\theta+\Delta \theta$$ and $$z$$ and $$z+\Delta z$$ and claculate its area.

CB
Actually, I did draw a picture, still didn't help.

Thanks anyway.
Mac

##### Member
First observe that you have a cone with slant height $$2 \sqrt{5}$$ and base radius $$2$$, so the curved surface area is the circumference of the base ($$4 \pi$$) times the slant heigh over two = $$4 \pi \sqrt{5}$$.

Now lets do this using integration. First switch to cylindrical polars, then the surface element is $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$, so the area is:

$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz$

etc...

(you should always draw a picture for these problems if you can)
Just a quick observation,

$$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz \neq 4 \pi \sqrt{5}$$

Last edited:

#### dwsmith

##### Well-known member
Just a quick observation,

$$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz \neq 4 \pi \sqrt{5}$$
$$\int_0^{2\pi}d\theta = 2\pi\Rightarrow\int_0^4\pi\sqrt{5}zdz = \frac{\pi\sqrt{5}}{2}\left.z^2\right|_0^4 = \frac{16\pi\sqrt{5}}{2} = 8\pi\sqrt{5}$$
Are you missing a factor of 1/2 with the Jacobian or something?
In post 2, it shows two factors of 1/2 which would be 1/4. Making that correction, we would have the desired results.

Now lets do this using integration. First switch to cylindrical polars, then the surface element is $$(z/2)\;d\theta\; (\sqrt{5}/2) dz$$, so the area is:

$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz$

etc...

(you should always draw a picture for these problems if you can)

So I think the integral should have been expressed as

$$\int_{0}^4 \int_{0}^{2\pi} \frac{\sqrt{5}}{4} z d\theta dz$$

Last edited by a moderator:

##### Member
I have solved the integral the following way.

$$\int_0^{2\pi}\int_0^2 (\sqrt{5})rdrd\theta = 4\pi\sqrt{5}$$

Thanks everyone for their contributions.

Last edited:

#### CaptainBlack

##### Well-known member
Just a quick observation,

$$A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz \neq 4 \pi \sqrt{5}$$
There is a typo that is obvious if you look at the previous line, that is supposed to be the integral of the area element which has an extra 1/2 in it that has got lost.

Anyway it is fixed now.

CB

Diagram: