Multivariable integration problem

[danny_manny]

Homework Statement

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)‖ dtdsFind 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:

Homework Equations

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.

thanks for your help.

 

[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.

 

[danny_manny]

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

 

[SammyS]

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

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.

 

[danny_manny]

Ok, thanks Sam.

Danny

 

[Dick]

Ok, thanks Sam.

Danny

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.

 

[danny_manny]

mmm I must have made a careless error.

Danny

 

[Dick]

mmm I must have made a careless error.

Danny

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

 

[danny_manny]

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

 

[Dick]

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

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

 

[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

 

[Dick]

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

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?

 

[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.

 

[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.

 

[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?

 

[Dick]

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.

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.

 

[danny_manny]

Ah! thanks for the help :)