Calculus 3 (Integral where z is in terms of two functions)

[Chantry]

Homework Statement

Find the mass and center of mass of the solid E with the given density function p.

38. E is bounded by the parabolic cylinder z = 1 - y^2 and the planes x + z = 1, x = 0, z = 0; p(x,y,z) = 4.

Homework Equations

The Attempt at a Solution

I've been trying to find a solution to this all morning.

I graphed it on paper in 3 dimensions and also graphed the zx, zy, and xy planes. The zx plane had equation z = 1-x, the zy had z = 1 - y^2, and the xy had two lines y = 1 and y = -1.

The bounds that I came up with were {-1 <= y <= 1, 0 <= x <= 1, 1-x<=z<=1-y^2}

This is where I had difficulty. The bounds I reasoned for z came from not being able to directly compare the equations z = 1-y^2 and x + z = 1. I'm not sure if I'm allowed to bound one of the axis in integration with two functions of unrelated variables. I guess you could argue that x and y are in both equations, except they're zero, but I have a feeling that's wrong.

Anyway, I solved the integral:
$$\int_{0}^{1}\int_{0}^{1}\int_{1-x}^{1-y^{2}}4dzdxdy$$

Which came out to 16/3. This seems reasonable, but I thought I'd see what you guys had to say about it.

I know how to find the center of mass. I just don't know if I'm setting up the integral correctly.

Any help would be greatly appreciated :).

Chantry

 

[LCKurtz]

Homework Statement

Find the mass and center of mass of the solid E with the given density function p.

38. E is bounded by the parabolic cylinder z = 1 - y^2 and the planes x + z = 1, x = 0, z = 0; p(x,y,z) = 4.

Homework Equations

The Attempt at a Solution

I've been trying to find a solution to this all morning.

I graphed it on paper in 3 dimensions and also graphed the zx, zy, and xy planes. The zx plane had equation z = 1-x, the zy had z = 1 - y^2, and the xy had two lines y = 1 and y = -1.

The bounds that I came up with were {-1 <= y <= 1, 0 <= x <= 1, 1-x<=z<=1-y^2}

This is where I had difficulty. The bounds I reasoned for z came from not being able to directly compare the equations z = 1-y^2 and x + z = 1. I'm not sure if I'm allowed to bound one of the axis in integration with two functions of unrelated variables. I guess you could argue that x and y are in both equations, except they're zero, but I have a feeling that's wrong...

While it is true that the xy domain is a square, choosing to integrate in the z direction first is a poor choice. The reason is that the "roof" of this solid consists of two separate parts; part of the roof is on the curved parabola shape and part is on the slanted plane. So if you wish to integrate in the z direction first you have to do two separate integrals. The best thing is to set it up again integrating either dx or dy first. You will see that one triple integral will work and the outer two limits will be easy from your picture.

And of course this integral only calculates the mass and you need an additional integral to do the center of mass, but you probably knew that.

 

[Chantry]

Thanks a lot. I really appreciate the help.

Alright, so I tried setting the bounds like this:

{-1 <= y <= 1, 1 - y^2 <= z <= 1, 0 <= x <= 1-z}

I integrated using 2 multiplied by the integral with bounds for y of 0 <= y <= 1,
and I got a result of 4/5.

I'm hoping I've finally wrapped my head around this.
Do those bounds sound right to you?

 

[LCKurtz]

Thanks a lot. I really appreciate the help.

Alright, so I tried setting the bounds like this:

{-1 <= y <= 1, 1 - y^2 <= z <= 1, 0 <= x <= 1-z}

I integrated using 2 multiplied by the integral with bounds for y of 0 <= y <= 1,
and I got a result of 4/5.

I'm hoping I've finally wrapped my head around this.
Do those bounds sound right to you?

I don't like your z limits. z is never larger than 1 - y².

 

[Chantry]

The way my drawing looks is I essentially have a -y^2 parabola starting at z = 1 on the yz plane.

So the way I thought about it was that on the z-axis it's at a maximum of 1, and then falls down on each side as y increases.

What's my error in thinking there?

 

[Chantry]

Nevermind. I get it.

It would be 0 <= z <= 1-y^2, because it's bounded by the equations z = 0, and z = 1-y^2.

Right?

 

[LCKurtz]

Nevermind. I get it.

It would be 0 <= z <= 1-y^2, because it's bounded by the equations z = 0, and z = 1-y^2.

Right?

Yes.

 

[Chantry]

Thank you! :)