How to solve a surface double integral?

[al_ex]

Homework Statement

surface double integral, sphere, cardioid

Relevant Equations

how do I plan the problem?

Hi I´d like a suggestion about a surface double integral. If I have a sphere x^2+y^2+z^2=4 is on the top of a cardioid r=1-cosθ. The problem is when I solve the integral I got a inverse sine when the answer is a natural logarithm (ln)

 

[LCKurtz]

Problem Statement: surface double integral, sphere, cardioid
Relevant Equations: how do I plan the problem?

Hi I´d like a suggestion about a surface double integral. If I have a sphere x^2+y^2+z^2=4 is on the top of a cardioid r=1-cosθ. The problem is when I solve the integral I got a inverse sine when the answer is a natural logarithm (ln)

Hi al_ex. Welcome to PF. You have to show some effort before anyone will help you. I would suggest you take a look at this insight article:
https://www.physicsforums.com/threa...zation-and-surface-integrals-comments.970135/Then try parameterizing your surface in cylindrical coordinates and come back if you get stuck. Also you might look at the LaTeX guide for typing equations:
https://www.physicsforums.com/help/latexhelp/

 

[al_ex]

I am parameterizing dz/dx and dz/dy I got this r/(4-r^2)^(1/2) dr dθ and my parameters are from 0 to 1-cosθ and from 0 to π but when I integrate r/(4-r^2)^(1/2) dr, I got (4-r^2)^(1/2) and when I put the values 0 and 1-cosθ is where I´m starting to get a troubles because I can´´ not get in the second integration the natural logarithm I think I am planning wrongly my equations. I really need help. The answer of this problem is 8 (π -(2)^1/2-ln<(2)^1/2+1>) I don't think that the book´'s answer is wronged.

 

[LCKurtz]

When I parameterize it with $r$ and $\theta$ I get$$
A=\int_0^{2\pi}\int_0^{1-\cos\theta}~\frac{2r}{\sqrt{4-r^2}}~drd\theta$$
for which Maple gives $8\pi -8\sqrt 2 +4 \ln(3-2\sqrt 2)$. I think your $\theta$ limits are wrong and anything else is difficult to guess without seeing your steps.

 

[LCKurtz]

Let me add...in a more direct response to your integration problem, while Maple easily evaluates $\int_0^{2\pi}\sqrt{4-(1-\cos\theta)^2}~d\theta$, it gives a horribly messy antiderivative. It's no wonder you are having trouble integrating it. Once you have it properly set up I wouldn't waste any more time trying to evaluate it unless you have an appropriate software program to do it for you.

 

[al_ex]

what do you suggest me? what can I do? this problem has any solution? can you give me a hint? please...

 

[LCKurtz]

In post #4 I have given you a correct setup and a correct answer to check your work against. Assuming you agree with that, you have the problem set up correctly. Like I said in post #5, actually working out the antiderivative looks to be complicated and time consuming and a waste of time. I don't plan to waste any more time trying to do it. Ask your teacher how he/she would proceed.

 

[vela]

While Maple easily evaluates $\int_0^{2\pi}\sqrt{4-(1-\cos\theta)^2}~d\theta$, it gives a horribly messy antiderivative.

It doesn't look too bad to do by hand starting with the half-angle trig identity. I got it down to an integral of the form
$$\int \sqrt{1+u^2}\,du.$$ You can look this last one up in a table of integrals.