Why multiply by 4 when setting up volume integral for elliptic paraboloid

[FeDeX_LaTeX]

Homework Statement

Find the volume of the region bounded by the elliptic paraboloid $$z = 4 - x^2 - \frac{1}{4}y^2$$ 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:

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

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.

 

[SteamKing]

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.

 

[FeDeX_LaTeX]

Ah that makes a lot more sense, thanks!