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Double Integral Help

  • #1
FeDeX_LaTeX
Gold Member
437
13

Homework Statement


Find the volume of the region bounded by the elliptic paraboloid [tex]z = 4 - x^2 - \frac{1}{4}y^2[/tex] and the plane z = 0.

Homework Equations


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The Attempt at a Solution


I'm not really sure where to start with this. This is how they've set it up:

[tex]4 \int_{0}^{2} \int_{0}^{2 \sqrt{4 - x^2}} \left( 4 - x^2 - \frac{1}{4}y^2 \right) dy dx[/tex]

Looking at the graph hasn't helped me understand how they got this. How did they set the integral up in this way?

I can see that they've got that upper limit of 2*sqrt(4 - x^2) by letting z = 0 and finding y in terms of x. But I haven't the faintest idea why they're integrating from 0 to 2 next, nor why they are multiplying the whole thing by 4... any help?

I would guess that the multiplying by 4 is due to the symmetry of the surface, but I don't understand anything else.
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
The 4 in front of the double integral tells you that the integral itself covers one quadrant of the total volume.

You are given a function f(x,y) = z and are told to find the volume occupied between the plane z = 0 (which is the x-y plane) and f(x,y) = 0. This is analogous in 2-D geometry to finding the area under a parabola y = x^2 and y = 0 between two values of x.

The problem has been pre-digested for your convenience. All you have to do is turn the crank on evaluating the double integral.

To find the range of x values for the elliptic paraboloid, set y and z = 0 and solve for x.

As always, a sketch can illuminate greatly.

The problem has been pre-digested for your convenience. All you have to dois turn the crank on evaluating the double integral.
 
Last edited:
  • #3
FeDeX_LaTeX
Gold Member
437
13
Ah that makes a lot more sense, thanks!
 

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