Help me integrate this volume please

[zeus77]

Homework Statement

Volume enclosed by: z= x^2 + y^2 and z= 2y.
This should look like 1/2 of a bowl only on the +y side of the x y z plane I believe. I want to know if integrating in different order or cord system will affect the result. I am having a lot of problems attempting to find the answer to this part of the equation:

V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy] = Pi/2 ?

My teacher says he found the volume to be Pi/2. Which i think might be incorrect.

I need to show the process of integration of dydzdx and dxdydz and compare the results. Or change to spherical coordinates.

2. The attempt at a solution

I am having an horrendous time finding the correct bounds for this i believe is my problem. Along with the fact i do not understand how to use my accessible integration tables to correctly solve the last part.

Starting with:

V= Int[0->2]dy Int[y^2 -> 2y]dz Int [-sqrt(z-y^2) -> sqrt(z-y^2)]

I move down correctly to,

V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy]

V= Pi/2 ?

Can some one please explain to me the correct integration table substitution of this part? Every table i look at has things in the form of:

Int[sqrt(2au-u^2)du , Int[sqrt(u^2-a^2) or Int[sqrt(a^2-u^2)

How do you convert my problem which seems to have the form of Int[{(sqrt(2au-u^2))^3}du] ?? I've tried of substitution and i feel uncertain of the result. If someone could explain how that integral equals Pi/2 i should be able to change to spherical cords or integrate in a different order.

Thanks in advance!

 

[zeus77]

I would solve this using Mathematica on my computer, but for some reason it rolled back my registration and i can't access it. I would like to have some work done for this by tomorrow, and i know that my professor does not like when we "cheat" using a computer to do integrals anyways hahaha.

 

[PingPong]

Your professor's answer is correct.

To start you off, for your integral, [itex]z[/itex] is going from [itex]x^2+y^2[/itex] to [itex]2y[/itex]. Now you're left with the image of a circle in the x-y plane. What should your limits of integration be there?

You could try a different coordinate system as well - since the projection in the x-y plane is a circle, what do you think would be best?

 

[zeus77]

Thanks for fast response pingpong.

Ok, so how did you figure his answer was correct? Were you able to use an integral table? cause i would really like to know that process... i have no clue on how to do
V= Int(0->2) [4/3 (2y-y^2)^(3/2) dy]= Pi/2 . without a computer's help that is.

I'm still confused on the limits of x and y also... Wont the projection onto the x/y plane be just 1/2 of a circle? Since the values of -y won't be included in the volume?

I would like to try to use spherical coordinates but I am just so uncertain of how to set it up, it seems like every example I've done is much simpler... All 3 calc books i have don't give me any good examples, or I'm not quick enough to understand them.

 

[zeus77]

Just getting frustrated on this problem now... i feel that I've lost understanding on how to set up the limits let alone understand how the integral is going to work.

Any help would be amazing

 

[zeus77]

So it's been a couple days so far, and I've still not been able to change the cords of this problem to spherical or cylindrical and reproduce my answer of pi/2...

Can some one look at my bounds for when i tried to change to Cylindrical Cords?

Original problem restated----Volume bounded by z= x^2 + y^2 and z=2y.

Cylindrical Cords attempt.
Bounds of Z, theta and r.
r^2<z<2rsin(theta) -pi < theta < pi 0<r<2

V= (Triple integral with bounds above) r^2 cos(theta) dzdrd(theta)
Which i evaluate to an answer not equal to pi/2.

 

[HallsofIvy]

So it's been a couple days so far, and I've still not been able to change the cords of this problem to spherical or cylindrical and reproduce my answer of pi/2...

Can some one look at my bounds for when i tried to change to Cylindrical Cords?

Original problem restated----Volume bounded by z= x^2 + y^2 and z=2y.

Cylindrical Cords attempt.
Bounds of Z, theta and r.
r^2<z<2rsin(theta) -pi < theta < pi 0<r<2

V= (Triple integral with bounds above) r^2 cos(theta) dzdrd(theta)
Which i evaluate to an answer not equal to pi/2.

Where did the "r^2 cos(theta)" come from? That looks like spherical coordinates. The differential for cylindrical coordinates is r dz dr d(theta).

 

[zeus77]

Thats a good point halls of ivy... some how i manage to combine the spherical and change it to cos instead of sin. Let me attempt it with that change!