Multivariable Calculus - Integration Assignment

[ConnorM]

Ok I see how it works now thank you. So then my integral would be,

abc(a^2 +b^2)ρ∫V′′r4sin^3(ϕ)dV′′

This has both of the Jacobians we found, from here I would integrate to find my answer then? My bounds of integration would have to be 0 ≤ θ ≤ 2pi , 0 ≤ ϕ ≤ pi , 0 ≤ r ≤ 1 ?

 

[STEMucator]

Ok I see how it works now thank you. So then my integral would be,

abc(a^2 +b^2)ρ∫V′′r4sin^3(ϕ)dV′′

This has both of the Jacobians we found, from here I would integrate to find my answer then? My bounds of integration would have to be 0 ≤ θ ≤ 2pi , 0 ≤ ϕ ≤ pi , 0 ≤ r ≤ 1 ?

Somehow the limits look good, but the integral looks wrong. I already showed you what you should wind up integrating a few posts ago. What do you get when integrating this?

Some advice: Work on determinants/trig/algebra a little bit to make this sort of thing easier in the future.

 

[ConnorM]

Ok I think I messed up and thought that I could factor out the sin^2θ + cos^2θ to simplify to 1. I will try again.

 

[ConnorM]

So now after integrating abcρ∫V′′(a2cos2(θ)+b2sin2(θ))r4sin3(ϕ)dV′′ , I have

(8pi*a^3 bcρ) / 15

 

[STEMucator]

So now after integrating abcρ∫V′′(a2cos2(θ)+b2sin2(θ))r4sin3(ϕ)dV′′ , I have

(8pi*a^3 bcρ) / 15

You have an iterated integral when you plug in the limits, right? So it doesn't matter too much what order of integration you decide to use.

Usually, you would want to pick the convenient one first. Either $\phi$ or $r$ in this case.

Check your answer again though.

 

[ConnorM]

This time I got, (3pi^2 r^4 (a^2 + b^2)abcρ)/8

 

[STEMucator]

This time I got, (3pi^2 r^4 (a^2 + b^2)abcρ)/8

Still not quite. I'm not sure where you are making the error, but it's in there. As a hint, what is the result of doing the first integration inside the big brackets:

$$abc \rho \int_0^1 \int_0^{2\pi} (a^2 cos^2(\theta) + b^2 sin^2(\theta)) r^4 \left[ \int_0^{\pi} sin^3(\phi) \space d \phi \right] d \theta d r$$

 

[ConnorM]

Doing the first integration I found that,

1/12 (cos(3ϕ) - 9cos(ϕ))

then subbing in ϕ = pi and ϕ = 0

1/12 (cos(3pi) - 9cos(pi)) - 1/12 (cos(0) - 9cos(0)) = 2/3 - (-2/3) = 4/3

 

[ConnorM]

After that I can integrate for dr,

(a2cos2(θ)+b2sin2(θ))r^5 /5

then subbing in r = 1 and r = 0

(a^2cos^2(θ)+b^2sin^2(θ))/5 - 0 = (a^2cos^2(θ)+b^2sin^2(θ))/5

 

[STEMucator]

Doing the first integration I found that,

1/12 (cos(3ϕ) - 9cos(ϕ))

then subbing in ϕ = pi and ϕ = 0

1/12 (cos(3pi) - 9cos(pi)) - 1/12 (cos(0) - 9cos(0)) = 2/3 - (-2/3) = 4/3

Okay, so now what happens here:

$$abc \rho \int_0^1 \frac{4}{3} r^4 \left[ \int_0^{2\pi} (a^2 cos^2(\theta) + b^2 sin^2(\theta)) d \theta \right] d r$$

Hint: You need to integrate $cos^2(\theta)$ and $sin^2(\theta)$ separately, treating $a^2$ and $b^2$ as constants.

 

[ConnorM]

Then I can integrate for dθ,

(a^2 θ)/2 + (a^2 sin(2θ)) /4 + (b^2 θ)/2 - (b^2 sin(2θ)) /4

Subbing in 2pi and 0,

(a^2 2pi)/2 + (a^2 sin(4pi)) /4 + (b^2 2pi)/2 - (b^2 sin(4pi)) /4 - ( (a^2 0)/2 + (a^2 sin(0)) /4 + (b^2 0)/2 - (b^2 sin(0)) /4 )

= a^2 pi + b^2 pi

So my final answer is pi(a^2 + b^2)

 

[STEMucator]

Then I can integrate for dθ,

(a^2 θ)/2 + (a^2 sin(2θ)) /4 + (b^2 θ)/2 - (b^2 sin(2θ)) /4

Subbing in 2pi and 0,

(a^2 2pi)/2 + (a^2 sin(4pi)) /4 + (b^2 2pi)/2 - (b^2 sin(4pi)) /4 - ( (a^2 0)/2 + (a^2 sin(0)) /4 + (b^2 0)/2 - (b^2 sin(0)) /4 )

= a^2 pi + b^2 pi

So my final answer is pi(a^2 + b^2)

Right, now just clean up the last integration, what do you get?

 

[ConnorM]

Oh by the way is there a tutorial or like a forum post somewhere on how to format the equations like you do? Makes it a lot easier to read, sorry for typing it out like I am!

 

[ConnorM]

Oops forgot about abcρ that was outside, so the answer is

abcρ * pi (a^2 + b^2)

 

[STEMucator]

Why not relax and carefully write the steps out as you did before? Your answer seems to be missing a factor of $\frac{4}{15}$. Anytime you rush math and forget to write something down, you're going to make a mistake like that.

Also, a useful LaTeX reference is located here:

https://www.physicsforums.com/showthread.php?p=3977517#post3977517

 

[ConnorM]

Ok I see now, yes I was rushing through things quickly and forgot that I had taken out the 4/3 from integrating dϕ and I had taken out 1/5 after integrating dθ. So that gives me the factor of 4/15.

final answer is now

(4/15) *abcρ * pi (a^2 + b^2)

 

[STEMucator]

Ok I see now, yes I was rushing through things quickly and forgot that I had taken out the 4/3 from integrating dϕ and I had taken out 1/5 after integrating dθ. So that gives me the factor of 4/15.

final answer is now

(4/15) *abcρ * pi (a^2 + b^2)

Good. That is indeed the answer and I'm glad you see why.

 

[ConnorM]

Thanks a lot, thank you for bearing with me all the way through!