# Multiple Integration Limits

1. Mar 24, 2009

### Sparky_

I'm throwing in the towel - I need some explanation help / clarification -

I'm brushing up on my polar / cylindrical coordinates and integration with polar / cylindrical coordinates.

In working through some examples - the book seems to write the limits as if it is straightforward or I should already know how to generate the limits - it just states them and moves on. (I can do the integration once I have the limits.)

With working backwards from the limits given I agree with the limits.

For example - an example states, "calculate the volume of the ellipsoid
$$4x^2 + 4y^2 + z^2 = 4$$

The limits given are:

$$0<= z <= 2\sqrt{(1-x^2 - y^2)}$$

I agree with this: solve for z

next

$$-\sqrt{(1-x^2)} <= y <= \sqrt{(1-x^2 )}$$

Not as sure about this - do we set z= 0, do we ignore z, if so why?

$$-1 <= x <= 1$$

Just like the limits on y, do I ignore z and y or set them equal to 0 , if so why?

Can you explain how to acquire these limits?

Thanks
Sparky

2. Mar 24, 2009

### HallsofIvy

Staff Emeritus
Because, having integrated with respect to z, you are now working in the xy-plane. It is not so much "setting z=0" as "projecting the figure onto the xy-plane". For an ellipse, projecting onto the xy-plane gives the outline of the ellipse which is the same as taking z= 0. If this were, say, an cone such as $z= \sqrt{x^2+ y^2}$ with z from 0 to 2, although taking z=0 gives the single point (0,0), projecting the entire ellipse to the xy-plane gives the circle $x^2+ y^2= 4$.

Because projecting every point in the ellipse onto the x-axis, you get every value of x from -1 to 1.

I prefer to think in the other direction: from the outside in: In order that triple integral be a number rather than a function of one or more variables, the limits of the outer integral must be a constant. Yet the integral must "cover" the entire figure. If that outer integral is with respect to, say, x, the limits must be the smallest and largest possible values of x in that figure.

In this case, the ellipse $$4x^2 + 4y^2 + z^2 = 4$$, which we could also write as $$\frac{x^2}{1}+ \frac{y^2}{1}+ \frac{z^2}{4}= 1$$, x ranges from -1 to 1, y from -1 to 1 and z from -2 to 2. If we decide to integrate so that the x-integral is the "outer" integral, its limits will be -1 to 1. If we decide to integrate so that the y-integral is the "outer" integral, its limits will be -1 to 1. If we decide to integrate so that the z-integral is the "outer" integral, its limits will be -2 to 2.

If, having set up the "outer integral" with respect to x, the limits of the next integral can depend on x but not on y or z. If we decide to do that integral with respect to y, we need to use the least and greatest values of y for a given x. Again, it is not just "taking z= 0" but projecting the figure onto the xy-plane. In this case the "widest" part of the ellipse happens to be when z= 0:
$$4x^2+ 4y^2= 4$$
so the maximum and minimum y are $\pm\sqrt{1- x^2}$.

For given x and y, the boundary is given by solving $$4x^2+ 4y^2+ z^2= 4$$ solving that for z gives $$z= \pm\sqrt{4- 4x^2- 4y^2}$$.

Lets look at the problem I mentioned before: find the volume of the cone $$z= \sqrt{x^2+ y^2}$$ between z= 0 and z= 2. The "broadest" part of that is at the top:z= 2 where $$2= \sqrt{x^2+ y^2}$$ or $$x^2+ y^2= 4$$, the circle centered on the origin with radius 2. every point in the cone projects down to a point in that circle. Projecting that circle to the x-axis gives the interval from x=-2 to x= 2.

The limits on the "outer" integral, with respect to x would be -2 and 2. On that circle, drawing a vertical line for a given x, y ranges from $$-\sqrt{4- x^2}$$ to $$\sqrt{4- x^2}$$. Those are the limits of integration on the "dy" integral.

Finally, the part of the cone that we want is above $$z= \sqrt{x^2+ y^2}$$ and below z= 2. The integral for the volume of the cone is
$$\int_{x= -2}^2\int_{y= -\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{x^2+ y^2}^2 dzdydx$$

3. Mar 27, 2009

### Sparky_

HallsofIvy

Thank you so much!!