Multivariable Calculus - Integration Assignment

[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!