Finding Bounds for a Triple Integral: Solids Bounded by a Paraboloid and Plane

[harpazo]

Set up a triple integral for the volume of the solid. DO NOT FIND THE VOLUME.

The solid bounded by the paraboloid z = 9 - x^2 - y^2 and the plane z = 0.

We are dealing with the xy-plane where z = 0.

I know that 0 ≤ z ≤ 9 - x^2 - y^2. The bounds of the integral pertaining to dz are z = 0 to z = 9 - x^2 - y^2.

How do I find the other bounds?

 

[MarkFL]

$z$ is going to range from $z=0$ to $z=9$. Along any of these planes, the cross-section will be the circle:

\(\displaystyle x^2+y^2=9-z\)

And so $y$ will range from $y=-\sqrt{9-z}$ to $y=\sqrt{9-z}$

And $x$ will range from $x=-\sqrt{9-z-y^2}$ to $x=\sqrt{9-z-y^2}$

Thus, the volume will be given by:

\(\displaystyle V=\int_0^9\int_{-\sqrt{9-z}}^{\sqrt{9-z}}\int_{-\sqrt{9-z-y^2}}^{\sqrt{9-z-y^2}}\,dx\,dy\,dz\)

If we use polar coordinates, then we have:

\(\displaystyle 0\le\theta\le2\pi\)

\(\displaystyle 0\le r\le3\)

\(\displaystyle 0\le z\le9-r^2\)

And the volume is:

\(\displaystyle V=\int_0^{2\pi}\int_0^3 r\int_0^{9-r^2}\,dz\,dr\,d\theta\)

 

[MarkFL]

Just so you have something to check your answers against, observe that we can find the volume using a solid of revolution, a Calc I technique.

\(\displaystyle V=\pi\int_0^9 x\,dx=\frac{81\pi}{2}\) :D

 

[harpazo]

I recall very little about the solid of revolution from calculus 1.

 

[Prove It]

I recall very little about the solid of revolution from calculus 1.

Then you will need to go back and review them!

 

[harpazo]

I need to review solids of revolution from calculus 1.