How to solve a surface double integral?

In summary: Hope that helps. In summary, the problem is set up correctly but the antiderivative is complicated and time consuming to evaluate.
  • #1
al_ex
3
0
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)
 
Physics news on Phys.org
  • #2
al_ex said:
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/
 
  • #3
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.
 

Attachments

  • 1556817934815.png
    1556817934815.png
    344 bytes · Views: 169
  • #4
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.
 
  • #5
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.
 
  • #6
what do you suggest me? what can I do? this problem has any solution? can you give me a hint? please...
 
  • #7
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.
 
  • #8
LCKurtz said:
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.
 

Related to How to solve a surface double integral?

1. What is a surface double integral?

A surface double integral is a type of integral used in multivariable calculus to calculate the volume under a curved surface in three-dimensional space. It is represented by a double integral sign and involves integrating over two variables, usually x and y, to find the volume.

2. How do I set up a surface double integral?

To set up a surface double integral, you first need to determine the limits of integration for both variables. This can be done by graphing the surface and identifying the boundaries. Then, you need to determine the integrand, which is the function being integrated. This is usually given in the problem or can be calculated from the given surface equation. Finally, you can set up the integral using the limits of integration and the integrand.

3. What is the difference between a surface double integral and a regular double integral?

A regular double integral is used to find the area under a curve in a two-dimensional plane, while a surface double integral is used to find the volume under a curved surface in three-dimensional space. The main difference is that a surface double integral involves integrating over two variables, while a regular double integral only involves one variable.

4. What are some common techniques for solving surface double integrals?

Some common techniques for solving surface double integrals include using the Cartesian coordinate system, changing the order of integration, and using polar, cylindrical, or spherical coordinates. It is also helpful to visualize the surface and identify symmetries or patterns to simplify the integral.

5. How do I know if I have set up and solved a surface double integral correctly?

To check if you have set up and solved a surface double integral correctly, you can use the properties of integrals, such as linearity and symmetry, to verify your answer. You can also use software or online calculators to graph the surface and compare your calculated volume to the visual representation. Additionally, you can double check your calculations and make sure you have used the correct limits of integration and integrand.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
981
  • Calculus and Beyond Homework Help
Replies
2
Views
369
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
908
  • Calculus and Beyond Homework Help
Replies
9
Views
385
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
493
Replies
4
Views
979
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top