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!!