Finding the volume of the solid using change of variables technique

Win_odd Dhamnekar

Win_odd Dhamnekar

Junior Member
Joined
Aug 14, 2018
Messages
207
Find the volume of the solid bounded by the \(\displaystyle z = x^2 +y^2\) and \(\displaystyle z^2 = 4(x^2+y^2)\)

My attempt to answer this question:

1654493917190.png

\(\displaystyle \displaystyle\int_0^{2\pi}\displaystyle\int_0^2\displaystyle\int_0^4 1\rho^2\sin{(\phi)} d\rho d\phi d\theta =-\frac{128\pi\cos{(2)}-128\pi}{3}= 189.822143919 \)

Is this answer correct?
 
Last edited:
My answer is very different from yours. How did you arrive at your integral? What is [imath]\phi[/imath]? Is it limited to the [imath][0,2][/imath] interval? If you provide more details we'd have a better chance to help you.
 
blamocur said:
My answer is very different from yours. How did you arrive at your integral? What is [imath]\phi[/imath]? Is it limited to the [imath][0,2][/imath] interval? If you provide more details we'd have a better chance to help you.
Click to expand...
Are the following integrals correct? Did you compute the same answer as below?
\(\displaystyle \displaystyle\int_0^{2\pi}\displaystyle\int_0^2\displaystyle\int_0^4 r dz dr d\theta = 16\pi\)
 
It looks better to me: I like that you changed to [imath]r[/imath] and [imath]z[/imath] variables. But are you sure about you integration limits for [imath]z[/imath]?
 
blamocur said:
It looks better to me: I like that you changed to [imath]r[/imath] and [imath]z[/imath] variables. But are you sure about you integration limits for [imath]z[/imath]?
Click to expand...
I think integration limits for z should be -4 to 4. So, the answer would be \(\displaystyle 32\pi\). Do have I now correct answer?
 
Dhamnekar Winod said:
I think integration limits for z should be -4 to 4. So, the answer would be \(\displaystyle 32\pi\). Do have I now correct answer?
Click to expand...
Your "volumes" are not symmetric about x-y plane. So you cannot simply multiply by 2.

Check it with Wolframalfa.com
 
Dhamnekar Winod said:
I think integration limits for z should be -4 to 4. So, the answer would be \(\displaystyle 32\pi\). Do have I now correct answer?
Click to expand...
When I look at a cross-section of the solid I see that the limits for [imath]z[/imath] depend on [imath]r[/imath] -- agree?fig1.png
 
If my understanding is correct it is the volume below the red cone but above the blue paraboloid.
 
I don't know how to interpret the highlighting, but if you look at the cross-section in post # 7 then the solid is the result of rotation of that cross section. You can also figure out the solid from the equations in your first post.
 
Cartesian Coordinate.

\(\displaystyle \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2 + y^2}^{2\sqrt{x^2 + y^2}} dz \ dy \ dx = 8.37758\)

Cylindrical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{r^2}^{2r} r \ dz \ dr \ d\theta = 8.37758\)

Spherical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758\)
 
I got the same numerical value, but I wonder if you teacher expects the answer expressed through constants like [imath]\pi[/imath].
 
nasi112 said:
Cartesian Coordinate.

\(\displaystyle \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2 + y^2}^{2\sqrt{x^2 + y^2}} dz \ dy \ dx = 8.37758\)

Cylindrical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{r^2}^{2r} r \ dz \ dr \ d\theta = 8.37758\)

Spherical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758\)
Click to expand...
Hi,
So, I think you computed this volume. Is that correct?




1654580637911.png
 
nasi112 said:
Cartesian Coordinate.

\(\displaystyle \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2 + y^2}^{2\sqrt{x^2 + y^2}} dz \ dy \ dx = 8.37758\)

Cylindrical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\int_{r^2}^{2r} r \ dz \ dr \ d\theta = 8.37758\)

Spherical Coordinate.

\(\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758\)
Click to expand...
Hi,
1)Can we separately compute the volume V1 inside the paraboloid \(\displaystyle z =x^2 +y^2\) for 0 ≤ z ≤ 4?

2) Can we separately compute the volume V2 inside the cone \(\displaystyle z = 2\sqrt{x^2 +y^2}\) for 0 ≤ z ≤ 4?

3)Then can we deduct V1 from V2 to answer this question?

Examples of computation of volume of a paraboloid and a cone is given below:

1654584262765.png

1654584282857.png
 
Dhamnekar Winod said:
Hi,
1)Can we separately compute the volume V1 inside the paraboloid \(\displaystyle z =x^2 +y^2\) for 0 ≤ z ≤ 4?

2) Can we separately compute the volume V2 inside the cone \(\displaystyle z = 2\sqrt{x^2 +y^2}\) for 0 ≤ z ≤ 4?

3)Then can we deduct V1 from V2 to answer this question?

Examples of computation of volume of a paraboloid and a cone is given below:

View attachment 32962

View attachment 32963
Click to expand...
Yes, this is one way to do it. Personally, I found explicit integration of the difference using cylindrical coordinates the easiest way to go.
 
You must log in or register to reply here.
Top