Surface integral over an annulus

[bobred]

Homework Statement

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

Homework Equations

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

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

The Attempt at a Solution

Hi, would just like to know if what I have done is correct.

Integrating with respect to $$r$$

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

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

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

Does this look ok?

 

[lanedance]

effect, you've performed the integral
$$\int \int z(r,\theta) dr d \theta$$

i'm not really sure what that represents, you need to find an area element dA and sum up all the elements over the r, theta range

note that an area element changes with coordinate representation, for example for a flat area element
$$dA = dxdy = rdr d \theta$$

if you performed
$$\int \int z(r,\theta) dA = \int \int z(r,\theta) r dr d \theta$$

that would give you the volume of the surface above the xy plane

 

[lanedance]

see here on how to set up a surface area integral

http://en.wikipedia.org/wiki/Surface_integral

 

[lanedance]

updated post #2

 

[bobred]

So am I right in thinking I need to workout

$$\int \int \sqrt{(\frac{\partial}{\partial r})^2+(\frac{\partial}{\partial \theta})^2+1} r dr d \theta$$

James

 

[Dick]

So am I right in thinking I need to workout

$$\int \int \sqrt{(\frac{\partial}{\partial r})^2+(\frac{\partial}{\partial \theta})^2+1} r dr d \theta$$

James

Not really. That's not the form given in the reference lanedance gave. The derivatives are with respect to the wrong variables.

 

[hunt_mat]

I compute the answer as 5\pi /8

Mat

 

[jackmell]

Homework Statement

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

Homework Equations

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

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

The Attempt at a Solution

Hi, would just like to know if what I have done is correct.

Integrating with respect to $$r$$

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

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

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

Does this look ok?

That's kinda' confusing Bob. It's just the area of a surface right? If so, then first start with just the formula for the surface area of the function z=f(x,y) over some region R:

$$A=\int\int_R \sqrt{1+z_x^2+z_y^2}\;dxdy$$

See, that's nice and clean and no one would complain. You can do those derivatives right?

You end up with a very clean integral:

$$\int\int_R \sqrt{1+8y^2+8x^2}\; dxdy$$

How convenient is that! Now switch to polar coordinates with r^2=x^2+y^2 and that dxdy=rdrdt. Can you now figure the integration limits in polar coordinates?

 

[bobred]

Got there in the end

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