Double Integration Intervals: How to Find the Limits?

[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.

 

[Hurkyl]

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


For the lemniscate, I suggest sketching its graph.

 

[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.

 

[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..

 

[Hurkyl]

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

 

[HallsofIvy]

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.

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.

So since y is negative when x=3, the max for x is 2?

That's right.

Why is x allowed to be negative and not be bound by x=0?

'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.

 

[meister]

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

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.

 

[Hurkyl]

Using -?/4 to ?/4 as the interval, which is one lobe, I get 2, which is half the right answer.

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.