Surface integral over an annulus

In summary, the homework statement states that 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. Integrating with respect to r, the equation gives the following: 55/256, 55/128, 55/256, and 5/8. The attempt at a solution is to integrate with respect to r, r^
  • #1
bobred
173
0

Homework Statement



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

Homework Equations



The equation in polar form is [tex]r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2[/tex]

[tex]\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[/tex]

The Attempt at a Solution


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

Integrating with respect to [tex]r[/tex]

[tex]{\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
[/tex]

and with respect to [tex]\theta[/tex], I get the area as

[tex]\frac{5}{4}\pi[/tex]

Does this look ok?
 
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  • #2
effect, you've performed the integral
[tex]\int \int z(r,\theta) dr d \theta[/tex]

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
[tex]dA = dxdy = rdr d \theta[/tex]

if you performed
[tex]\int \int z(r,\theta) dA = \int \int z(r,\theta) r dr d \theta[/tex]

that would give you the volume of the surface above the xy plane
 
Last edited:
  • #4
updated post #2
 
  • #5
So am I right in thinking I need to workout

[tex]\int \int \sqrt{(\frac{\partial}{\partial r})^2+(\frac{\partial}{\partial \theta})^2+1} r dr d \theta[/tex]

James
 
  • #6
bobred said:
So am I right in thinking I need to workout

[tex]\int \int \sqrt{(\frac{\partial}{\partial r})^2+(\frac{\partial}{\partial \theta})^2+1} r dr d \theta[/tex]

James

Not really. That's not the form given in the reference lanedance gave. The derivatives are with respect to the wrong variables.
 
  • #7
I compute the answer as 5\pi /8

Mat
 
  • #8
bobred said:

Homework Statement



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

Homework Equations



The equation in polar form is [tex]r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2[/tex]

[tex]\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[/tex]

The Attempt at a Solution


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

Integrating with respect to [tex]r[/tex]

[tex]{\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
[/tex]

and with respect to [tex]\theta[/tex], I get the area as

[tex]\frac{5}{4}\pi[/tex]

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:

[tex]A=\int\int_R \sqrt{1+z_x^2+z_y^2}\;dxdy[/tex]

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:

[tex]\int\int_R \sqrt{1+8y^2+8x^2}\; dxdy[/tex]

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?
 
  • #9
Got there in the end

[tex]\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}[/tex]
 

FAQ: Surface integral over an annulus

What is a surface integral over an annulus?

A surface integral over an annulus is a type of multiple integral used to calculate the total surface area of a three-dimensional shape known as an annulus. It involves integrating a function over a surface that is bounded by two circles with different radii.

How is a surface integral over an annulus different from a regular surface integral?

A surface integral over an annulus is different from a regular surface integral because it is calculated over a specific type of surface, namely an annulus. Regular surface integrals can be calculated over any type of surface, such as a plane or a curved surface.

What is the formula for calculating a surface integral over an annulus?

The formula for calculating a surface integral over an annulus is ∫∫S F(x,y,z) dS = ∫∫D F(x,y,z(x,y))√(1 + ƒx2 + ƒy2) dA, where S is the surface of the annulus, D is the region bounded by the two circles, and ƒx and ƒy are the partial derivatives of the parametric equations for the surface.

What are some real-world applications of a surface integral over an annulus?

A surface integral over an annulus has many real-world applications, such as calculating the surface area of a cylindrical tank, determining the flux of a vector field through a cylindrical surface, or finding the total heat transfer from a circular pipe.

How is a surface integral over an annulus related to other types of integrals?

A surface integral over an annulus is related to other types of integrals, such as line integrals and volume integrals. It can be thought of as a combination of these two types of integrals, as it involves integrating over a curve (line integral) and a two-dimensional region (volume integral).

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