# Double integration Intervals

1. Sep 2, 2003

### meister

I'm currently taking multivariable calculus and we're studying double and triple integration. The actual integration itself is easy, but what I am not understanding at all is how to find the intervals for integration. I can only manage to find them on only the most simple problems. I don't need help solving the integral, just help finding the limits of integration. At least I hope I'm doing the integration correctly...

Example 1
Find the volume of the solid bounded by the cylinder x^2+z^2=9 and the planes x=0, y=0, z=0, x+2y=2 in the first octant.

For this problem, all variables must be positive, so, 0<=x<=3 and 0<=y<=1-x/2. This doesn't appear to yield the correct answer however...On this one I integrated [squ](9-x^2). Is that the correct function to integrate? The right answer for this one is (1/6)(11[squ]5-27)+(9/2)arcsin(2/3). I get something relatively close but ultimately incorrect.

Example 2
Use a double integral to find the area of the region enclosed by the lemniscate r^2=4cos(2theta).

On this I assumed the interval of r would be from 0 to when cos(2theta) was 1, so 0<=r<=4 and 0<=theta<=2pi. This doesn't yield the correct answer either, however. The right answer on this one is 4, I get 0.

Any help at all is greatly appreciated. Note that I don't need you to solve the problems for me, just help me understand how to compute the intervals. Thanks.

2. Sep 2, 2003

### Hurkyl

Staff Emeritus
For example 1, consider what the interval of allowed values for y is when x = 3...

For the lemniscate, I suggest sketching its graph.

3. Sep 2, 2003

### HallsofIvy

Your problem is that you are trying to use constants for both integrals: that's the same as integrating over a rectangle, not the figures you want.

1. Find the volume of the solid bounded by the cylinder x^2+z^2=9 and the planes x=0, y=0, z=0, x+2y=2 in the first octant.

Okay, geometrically, this is a quarter of the cylinder (with axis along the y-axis) bounded below by y=0 and above by y= 1- x/2. (Which meets y= 0 at x= 2 so does not actually include the whole cylinder: that has radius 3.)

Draw picture (I am extremely "drawing impaired" but I can draw "side views" involving only two axes at a time). Exactly what the limits of integration are depends upon the order of integration chosen. It's an interesting exercise to shift from one to another.

I think I will choose to write the volume as int(int (int dy)dz) dx
The "outer integral" is "dx" and so must have numerical limits: looking at my graph I see that x can range from x= -3 (the boundary of the cylinder) to x= 2 (where x+ 2y= 2 intersects y= 0).
For each value of x, z satisfies x^2+ z^2= 9 -> z^2= 9- x^2 or
z= +/- sqrt(9- x^2). In the "dz" integral, for each x, z can range from -sqrt(9- x^2) to +sqrt(9- x^2) and those are the limits of integration.
For the "inside integral- dy", the limits of integration can depend on both x and z. Here, because y only appears in the equations y=0 and x+ 2y= 2, y can range from 0 up to 1- x/2. Those are the limits of integration on the innermost integral.
You have int(x=-3 to 2){int(z=-sqrt(9-x^2) to sqrt(9-x^2){int(y= 0 to 1-x/2) dy} dz} dx.

2.Use a double integral to find the area of the region enclosed by the lemniscate r^2=4cos(2theta).
(Interesting figure. Assuming you want the area of both lobes, you need to let theta range from 0 to 2 pi.)
Here, of course, we will need to use polar coordinates. The "differential of area" in polar coordinates is r dr dtheta.

In order to get the entire figure we need to let theta range from 0 to 2 pi. (I mention that because, sometimes with that "2 theta" term, we find that just 0 to pi covers the figure. Going from 0 to 2pi covers twice and gives twice the area.)
FOR EACH THETA, r ranges from 0 out to the graph: r= sqrt(4cos(2theta)): notice we DON'T NEED r= - sqrt(4 cos(2theta)) in polar coordinates, r is always +, using negative values just reflects back the other way and we are already getting that.

The integral is int(theta= 0 to 2pi){int(r= 0 to sqrt(4cos(2theta)) r dr} dtheta.

4. Sep 2, 2003

### meister

Halls, on that first integral, what function do you integrate? For that particular problem I don't think we're supposed to use triple integration.

So since y is negative when x=3, the max for x is 2? Why is x allowed to be negative and not be bound by x=0?

Edit: I just changed the interval from -3<=x<=2 to 0<=x<=2 and got the right answer! Thanks a lot guys.

The second function is still puzzling me however..

5. Sep 2, 2003

### Hurkyl

Staff Emeritus
Did you try sketching the graph for problem two? The equation has a distinctive property... for example, consider what happens when &theta;=&pi;/4.

6. Sep 2, 2003

### HallsofIvy

The way I did it, since you asked for volume, there is no "function". Volume is the triple integral of dydzdx. Of course, you could go ahead and do the first integral: int(y= 0 to 1-x/2) dy
is 1-x/2 so the integral is int(x= 0 to 2) int(z= 0 to sqrt(9-x^2) dz dx.

That's right.
'Cuz I messed up! I was focusing on the "cylinder" and forgot about the "x= 0, y= 0, z=0" part! The integrals should be
int(x= 0 to 2) int(z= 0 to sqrt(9-x^2)) int(y=0 to 1-x/2) dydzdx.

7. Sep 2, 2003

### meister

It's the infinity sign, and at &pi;/4 the value of the function is zero..

Using -&pi;/4 to &pi;/4 as the interval, which is one lobe, I get 2, which is half the right answer. I can't seem to get it to be 4 however...if I make it 0 to 2&pi; I get a really small number approaching zero and using -&pi;/4 to 5&pi;/4 I get 2 also.

Last edited: Sep 2, 2003
8. Sep 3, 2003

### Hurkyl

Staff Emeritus
Both lobes should have the same area, right? So the total area is 4!

Like the other problem, you can't simply integrate from 0 to 2&pi; because some of those values are illegal. If you didn't want to use symmetry to determine the total area from knowing the area of a single lobe, I know of two ways to do it... (symmetry is still the best way, though):

(a) The total area is the sum of two integrals; one from -&pi;/4..&pi;/4 and the other from 3&pi;/4..5&pi;/4 (you can't integrate between &pi;/4 and 3&pi;/4 because the graph doesn't exist on that interval!)

(b) Instead of integrating the length from the origin to r for each &theta;, you could instead be integrating the length from -r to r, thus getting both lobes in your integral from -&pi;/4..&pi;/4.