Use polar coordinates to find the volume of the given solid.

[Julie323]

Homework Statement

Bounded by the paraboloid z = 4 + 2x2 + 2y2 and the plane z = 10 in the first octant.

Homework Equations

The Attempt at a Solution

Plugging in 10 for z I got 3=x²+y². From this, I set 0[tex]\leqr[/tex][tex]\leq3\sqrt{}[/tex].
I wasn't sure what to do with the first octant, but I guess that it meant 0[tex]\leq\vartheta[/tex][tex]\leq/2\pi[/tex]
I set my equation to be -6+2r²r dr d[tex]\vartheta[/tex].
After integrating I got -9[tex]\pi[/tex]/4
This was wrong. Any idea where I messed up? Thanks

 

[Julie323]

Wow, something messed up with the symbols, but I meant 0 less than or equal to r less than or equal to square root of 3

 

[vela]

Edit your original post and use LaTeX to typeset entire expressions instead of using it for just individual symbols. It'll be easier to read.

For the square root of 3, the code should be \sqrt{3}. What goes under the radical is inside the braces.

 

[Marioqwe]

First octant means that theta goes from 0 to pi/2.

 

[Julie323]

Basically my question is how to set it up. What I have right now is r going from 0 to \sqrt{3} and theta from 0 to pi/2 For my equation I have tried -6 +2r^2 r dr d(theta) and 4 + 2r^2 r dr d(theta). These gave me answers of -9pi/4 and 21pi/4. These are both wrong.
Does anyone see what I have set up wrong?

 

[vela]

Cleaned up your original post so it's readable.

Homework Statement

Bounded by the paraboloid z = 4 + 2x² + 2y² and the plane z = 10 in the first octant.

Homework Equations

The Attempt at a Solution

Plugging in 10 for z I got 3=x²+y². From this, I set [tex]0 \leq r \le \sqrt{3}[/tex].
I wasn't sure what to do with the first octant, but I guess that it meant [tex]0 \leq \vartheta \leq \pi/2[/tex]
I set my equation to be [tex](-6+2r^2)r dr d\vartheta[/tex].
After integrating I got [tex]-9\pi/4[/tex]
This was wrong. Any idea where I messed up? Thanks

The volume is given by

[tex]V=\int dV = \int (z_\mathrm{max}-z_\mathrm{min})\,dA[/tex]

For each little area element dA in the base, you multiply it by the height (z_max-z_min) to get an element of volume dV. In this case, you have an element of area [itex]dA=r\,dr\,d\vartheta[/itex], and you found the limits correctly. You just messed up the height.

It's always a good idea to draw a sketch to visualize what's going on. I've attached plots of the functions, a top view and a side view. Can you spot your mistake now?

 

[Julie323]

okay, so the equation I used also included the the portion of the paraboloid below the plane z=10? Do I need to subtract the portion below the plane? I don't know what to do with this kind of problem. I looked for an example like it in the book, but the only one was an even number so I do not have the solution to it. Thanks for the help!

 

[vela]

The volume is the portion below the plane. Imagine a vertical line running through the volume. What are z_max, the largest value of z, and z_min, the smallest value of z, on that line?

 

[Julie323]

z max would be 10 and z min would be 4. So I just integrate 4 r dr dtheta?

 

[vela]

No, that's not right. I should have been clearer. Both z_max and z_min are generally functions of r and θ. In this case, since the top is z=10, z_max=10 is a constant, but z_min will be the expression that describes the paraboloid.

 

[Julie323]

O okay. So the equation is 10-(4+2x²+2y²)? Which simplifies to 6-2r². Integrating this using my previous limits gives me 9pi/4. Does that look right? Thanks so much!

 

[vela]

Yup, that's the result I got too.