• Support PF! Buy your school textbooks, materials and every day products Here!

Volume of an ellipsoid using double integrals

  • Thread starter Lucas Mayr
  • Start date
  • #1
18
0

Homework Statement



Using double integrals, calculate the volume of the solid bound by the ellipsoid:

x²/a² + y²/b² + z²/c² = 1

2. Relevant data

must be done using double integrals

The Attempt at a Solution



i simply can't find a way to solve this by double integrals, i did with triple integral, but my teacher won't accept it, this is the last question and i can't find a solution for it, a step-by-step would be awsome.

Thanks.
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,856
1,654
If it were a solid of rotation you could do it in a single-integral right?
Basically you have to use reasoning in the place of one of the integrals ...

Find the volume between $$z=c^2\left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right )$$ and the x-y plane.
 
  • #3
34,370
10,446
i simply can't find a way to solve this by double integrals, i did with triple integral, but my teacher won't accept it, this is the last question and i can't find a solution for it, a step-by-step would be awsome.
The first integration in the triple integral is trivial, and afterwards you get a double-integral - you can use this double-integral (ignoring the first step) to solve the problem.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Essentially the same idea: projecting the ellipsoid to the xy-plane (z= 0) gives the ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex]. The two heights at each (x, y) point are [itex]z= \pm c\sqrt{1- x^2/a^2- y^2/b^2}[/itex]. The difference, [itex]2c\sqrt{1- x^2/a^2- y^2/b^2}[/itex], is the length of thin rectangle above that point and is the function to be integrated. For each x, The ellipse goes from [itex]y= -b\sqrt{1- x^2/a^2}[/itex] to [itex]y= b\sqrt{1- x^2/a^2}[/itex]. And, over all, x goes from -a to a. The volume of the ellipse is given by
[tex]\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}} 2c\sqrt{1- x^2/a^2- y^2/b^2} dydx[/tex].

As mfb says, this is the same as if you started with the triple integral
[tex]\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}}\int_{-c\sqrt{1- x^2/a^2- y^2/b^2}}^{c\sqrt{1- x^2/a^2- y^2/b^2}} dzdydx[/tex]
and integrated once.
 

Related Threads on Volume of an ellipsoid using double integrals

Replies
7
Views
1K
Replies
9
Views
8K
Replies
7
Views
12K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
7
Views
40K
  • Last Post
Replies
4
Views
866
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
33K
Replies
1
Views
3K
Replies
1
Views
4K
Top