Volume of solid bounded by 2 surfaces

  • Thread starter Kreamer
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  • #1
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Just working on some practice problems. I missed a couple classes due to sickness and just need some extra help. If you could walk me through how to do these types of problems that would be amazing.

Homework Statement
Evaluate the volume of the solid bounded by the surfaces
(x2 + y2)1/2 = z2
(x2 + y2)1/2 = 8 - z2
 

Answers and Replies

  • #2
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Hi, Kreamer. :)

These problems can be done with the use of double integrals.

What you're looking for is

[tex]V = \iint_\mathcal{D} (z_2(x, y) - z_1(x, y)) dxdy,[/tex]​

where [tex]z_2(x, y)[/tex] is the surface on top and [tex]z_1(x, y)[/tex] is the surface below. To get the region of integration you need to figure out where the two surfaces intersect and what this intersection looks like when projected onto the XY plane. To get a better idea, do a rough sketch of the situation.

When you construct the double integral, it may be hard or impossible to perform in cartesian coordinates, so you may need to do a change of variables - polar coordinates seems like an ideal candidate. ;) Don't forget the Jacobian!

Hope this helps.
 
  • #3
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I am starting to get it more. I am attempting the problem but am having trouble figuring out how exactly you get the right limits of integration.
 
  • #4
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(x2 + y2)1/2 = z2
(x2 + y2)1/2 = 8 − z2

I substituted then set z=0 and solved for x and y to get x = +/- (y2-16)1/2 and y = +/- (x2-16)1/2

Are these my limits or did I go wrong somewhere along the line?
 
  • #5
HallsofIvy
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(x2 + y2)1/2 = z2
(x2 + y2)1/2 = 8 − z2

I substituted then set z=0 and solved for x and y to get x = +/- (y2-16)1/2 and y = +/- (x2-16)1/2

Are these my limits or did I go wrong somewhere along the line?

Why did you set z= 0? If [itex](x^2+ y^2)^{1/2}= z[/itex] and [itex](x^2+ y^2)^{1/2}= 8- z[/itex], then z= 8- z so z= 4 where the two surfaces intersect. Now, when z= 4, (I think that was what you meant) both equations become [itex](x^2+ y^2)^{1/2}= 4[/itex] so that x^2+ y^2= 16. That is a circle, with center at the origin of the xy-plane and radius 4.

Remember than in a double integral, the limits of integration for the "outer" integral must be numbers so that your integral will be a number. The limits of integration for the "inner" integral can depend on the "other" variable.

That is, if you want to do the integral in the order
[tex]\int_x\int_y dy dx[/tex]
then the limits in the x-integral must be numbers while the limits for the y-integral may depend on x. To determine what those limits are, look at the circle. We want to "cover" the entire circle. On the left edge, x= -4 and on the right, x= 4 so those will be the limits for that integral. For each x y ranges between [itex]-\sqrt{16- x^2}[/itex] on the bottom to [itex]\sqrt{16- x^2}[/itex] on the top. Those will be the limits of integration for the y-integral:
[tex]\int_{x=-4}^4\int_{y= -\sqrt{16- x^2}}^{\sqrt{16- x^2}} f(x,y)dydx[/tex]

Of course, you could also take y from -4 to 4 and, for each y, x from [itex]-\sqrt{16- y^2}[/itex] to [itex]+\sqrt{16- x^2}[/itex], reversing the order of integration:
[tex]\int_{y= -4}^4\int_{x= -\sqrt{16- y^2}}^{\sqrt{16- y^2}} f(x,y) dxdy [/tex]

You probably haven't had double integrals in polar coordinates but, because of the circular symmetry here, that would give the simpest limits. You could take r from 0 to 4 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex].
 
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  • #6
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Ah yes yes I get it now. However I believe it is supposed to be 2, not 4. The equations were Z squared not just Z. But I think I get it now. Thank you!

Edit: Also I am not sure why I said set z=0, not what I did, just a bit tired I guess
 

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