Finding the volume of the solid using change of variables technique

[Win_odd Dhamnekar]

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:



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

Is this answer correct?

 

[blamocur]

My answer is very different from yours. How did you arrive at your integral? What is $\phi$? Is it limited to the $[0,2]$ interval? If you provide more details we'd have a better chance to help you.

 

[Win_odd Dhamnekar]

My answer is very different from yours. How did you arrive at your integral? What is $\phi$? Is it limited to the $[0,2]$ interval? If you provide more details we'd have a better chance to help you.

Are the following integrals correct? Did you compute the same answer as below?
$\displaystyle \displaystyle\int_0^{2\pi}\displaystyle\int_0^2\displaystyle\int_0^4 r dz dr d\theta = 16\pi$

 

[blamocur]

It looks better to me: I like that you changed to $r$ and $z$ variables. But are you sure about you integration limits for $z$?

 

[Win_odd Dhamnekar]

It looks better to me: I like that you changed to $r$ and $z$ variables. But are you sure about you integration limits for $z$?

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?

 

[Deleted member 4993]

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?

Your "volumes" are not symmetric about x-y plane. So you cannot simply multiply by 2.

Check it with Wolframalfa.com

 

[blamocur]

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?

When I look at a cross-section of the solid I see that the limits for $z$ depend on $r$ -- agree?

 

[Win_odd Dhamnekar]

When I look at a cross-section of the solid I see that the limits for $z$ depend on $r$ -- agree?View attachment 32954

Would you tell me which volume the questioner asked the readers to compute? Is it the volume of the solid given below excluding the blue part in upper half of xyz plane?

 

[blamocur]

If my understanding is correct it is the volume below the red cone but above the blue paraboloid.

 

[Win_odd Dhamnekar]

If my understanding is correct it is the volume below the red cone but above the blue paraboloid.

Would you mean to say the following highlighted volume?

 

[blamocur]

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.

 

[Deleted member 4993]

Would you mean to say the following highlighted volume?

View attachment 32956

You want to calculate the volume whose outer bound is the Orange cone and the inner bound is the green paraboloid (response #7). To my eye - your drawings are misleading. How did you draw those?

 

[nasi112]

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.

$\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758$

 

[blamocur]

I got the same numerical value, but I wonder if you teacher expects the answer expressed through constants like $\pi$.

 

[Win_odd Dhamnekar]

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.

$\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758$

Hi,
So, I think you computed this volume. Is that correct?

 

[Win_odd Dhamnekar]

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.

$\displaystyle \int_{0}^{2\pi}\int_{\tan^{-1}(1/2)}^{\pi/2}\int_{0}^{\cot\phi \csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta = 8.37758$

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:

 

[Win_odd Dhamnekar]

So, V₂=$\displaystyle \frac{32\pi}{3}$ and V₁=$\displaystyle 8\pi$.
Answer:
V₂ - V₁ = 8.37758

 

[blamocur]

Hi,
So, I think you computed this volume. Is that correct?




View attachment 32960

The numerical value is correct.

 

[blamocur]

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

Yes, this is one way to do it. Personally, I found explicit integration of the difference using cylindrical coordinates the easiest way to go.

 

[blamocur]

So, V₂=$\displaystyle \frac{32\pi}{3}$ and V₁=$\displaystyle 8\pi$.
Answer:
V₂ - V₁ = 8.37758

Yes, I got $\frac{8\pi}{3}$, too.