- #1

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**Homework Statement**

Evaluate the volume of the solid bounded by the surfaces

(x

^{2}+ y

^{2})

^{1/2}= z

^{2}

(x

^{2}+ y

^{2})

^{1/2}= 8 - z

^{2}

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- Thread starter Kreamer
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- #1

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Evaluate the volume of the solid bounded by the surfaces

(x

(x

- #2

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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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- #4

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(x

I substituted then set z=0 and solved for x and y to get x = +/- (y

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

- #5

HallsofIvy

Science Advisor

Homework Helper

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^{2}+ y^{2})^{1/2}= z^{2}

(x^{2}+ y^{2})^{1/2}= 8 − z^{2}

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

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.

[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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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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