# Homework Help: Multivariable integration problem

1. Feb 3, 2012

### danny_manny

1. The problem statement, all variables and given/known data

Vector calculus

It can be shown that the area of the surface described by the vector valued function
r(s; t) between the limits a ≤ s≤ b and c ≤ t ≤ d is given by

A=∫(from a to b) ∫(from c to d ) ‖(∂r/∂s)×(∂r/∂t)‖ dtds

Find the surface area of the bowl described by
r(s; t) = s cos(t)i + s sin(t)j + s^(2)k; 0 ≤ s ≤ 1; 0 ≤ t ≤ 2π:

2. Relevant equations

3. The attempt at a solution

Ok so first off I've solved this problem but am unsure if I am correct. The final answer I came to is roughly 50. However my friend thinks its roughly 5.33. I'm sure he is incorrect because in the last few steps the integral required a u substitution, where he didn't change the limits of integration in this case instead of 1 to 0 the new limits became 5 to 1.

if someone could check this I would be greatly appreciative.

Last edited: Feb 3, 2012
2. Feb 3, 2012

### danny_manny

Or is a change of the limits of integration even necessary in this problem? This is my only concern otherwise I'm confident my initial answer is correct.

3. Feb 3, 2012

### danny_manny

So if I use a u-substitution to solve a definite integral do i need to change the limits?

4. Feb 3, 2012

### SammyS

Staff Emeritus
If no one else checks this I'll try to get to it after I get home.

As for the question in the quotes text, YES, if you do a u-substitution to solve a definite integral, you should also change the limits of integration to be consistent with the substitution.

Alternatively, use u-substitution to find the anti-derivative, of course changing back to the original variable before evaluating the definite integral.

5. Feb 3, 2012

### danny_manny

Ok, thanks Sam.

Danny

6. Feb 3, 2012

### Dick

The norm of the two derivatives are both certainly not much more than 1. You are integrating over an area of 2*pi. 50 is certainly wrong. It's pretty easy to ballpark it that much.

7. Feb 4, 2012

### danny_manny

mmm I must have made a careless error.

Danny

8. Feb 4, 2012

### Dick

Doing it in detail, I agree with your friend's 5.33.

9. Feb 4, 2012

### danny_manny

Hey Dick, did you have to use a u-substitution for the last step to this problem?

10. Feb 4, 2012

### Dick

Sure I did. Used a trig identity too. What did you do? I think it's about time you told us.

11. Feb 5, 2012

### danny_manny

well the last step is where i seem to be going wrong i think. The second equation I get when i integrate is 2pi S(4S^(2)+1)^(1/2)

Which i then integrate using a u-substitution. still getting wrong answer

12. Feb 6, 2012

### Dick

That is the correct integral. But how are we supposed to know what is going wrong if you don't tell us what you got and how you got it?? I get around 5.33, your friend got around 5.33. Why don't you?

Last edited: Feb 6, 2012
13. Feb 6, 2012

### danny_manny

Ok so integrating that integral i set

u =4s^(2)+1
du=8s ds
(1/8)du = sds

subbing that into the original integral and integrating i get pi/6 (4s^(2)+1)^(3/2
and now i have to change my limits? From 1 to 0.

14. Feb 6, 2012

### danny_manny

so new limits will be

u=4(1)^(2)+1 = 5
and
u=4(0)^(2)+1 = 1

and using these new limits i get the wrong answer.

15. Feb 6, 2012

### danny_manny

however if i don't change the limits and keep them the same at 1 to 0 i get 5.33 :S, what am i missing?

16. Feb 6, 2012

### Dick

If you had written the answer in terms of u as pi/6 u^(3/2) then you would need to use the new u-limits. Since you changed the integral back to s, pi/6 (4s^2+1)^(3/2) you should use the original s limits, 0 to 1.

17. Feb 6, 2012

### danny_manny

Ah!! thanks for the help :)