# Help me integrate this volume please

• zeus77
In summary, the bounds for the volume are that r^2<z<2rsin(theta) and that 0<r<2. The integral with these bounds is V= (Triple integral with bounds above) r^2 cos(theta) dzdrd(theta). This function evaluates to an answer not equal to pi/2.
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.

Last edited:
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.

To start you off, for your integral, $z$ is going from $x^2+y^2$ to $2y$. 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?

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.

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

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.

Last edited:
zeus77 said:
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).

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!

## 1. How do I integrate a volume?

Integrating a volume involves finding the antiderivative of the function defining the volume and then plugging in the limits of integration. This will give you the volume of the region.

## 2. What is the purpose of integrating a volume?

Integrating a volume allows us to find the total amount of space enclosed by a given region. This is useful in a variety of scientific fields, such as physics and engineering.

## 3. What is the difference between integrating a volume and a surface area?

Integrating a volume involves finding the total space enclosed by a region, while integrating a surface area involves finding the total area of the surface of a region.

## 4. Can I use any integration method to find the volume of a region?

Yes, you can use various integration methods such as the disk method, shell method, or triple integrals to find the volume of a region. The choice of method depends on the shape of the region and the function defining it.

## 5. Are there any shortcuts or tricks to make integrating a volume easier?

There are some shortcuts and tricks, such as symmetry and using geometric properties to simplify the integration process. However, it is important to understand the underlying concepts and principles of integration to ensure accurate results.

• Calculus and Beyond Homework Help
Replies
20
Views
544
• Calculus and Beyond Homework Help
Replies
14
Views
414
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
825
• Calculus and Beyond Homework Help
Replies
4
Views
325
• Calculus and Beyond Homework Help
Replies
1
Views
535
• Calculus and Beyond Homework Help
Replies
8
Views
803
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
832
• Calculus and Beyond Homework Help
Replies
3
Views
393