Double Integral in Polar Coordinates

Up until now we have dealt with double integrals dealing with bounds in terms of x and y, known as cartesian coordinates. But what happens if this becomes relatively hard to do?

Consider the example of a circle, would you compute the area of a circle (if we didnt know that ) in cartesian coordinates? Probably not,

Using single integration we would probably find the area of the half hemisphere and multiply it by 2, that is

Assuming we didnt know the area that this integral represented, we would use the subsitution

However that's single integral thinking! And we now have double integrals by which to calculate these sorts of things. So how do we do that?

We tend to use cylindrical co-ordinates when there is a cylinder or sphere involved in the calculation.In polar co-ordinates we let

Noting that r denotes radius, we can easily compute the area of a circle.

If the above confuses you, don't worry! Let's go through it again.

Well...the first thing is to visualize. When we have a circle centered at the origin we can make a line from the origin (0,0) to any point on the circle of radius a. Now sweep that line, from (0,0) to a, all 360 degrees. You can even take your finger and do this. Notice how we've covered the entire circle! That's what double integration is, we are essentially sweeping over the area and by integrating with respect to bounded functions, we can obtain that area.

Consider the following problem

In carestian coordinates the above could be quite messy. I mean, we're looking at a lot of substitution! So instead, lets sub in polar coordinatesEvaluate: where D is the circle of radius a.

That means,

where Q is the domain of a circle of radius a.

Doesn't that look a lot more simple?! Clearly we find that

That was a simple example, and these problems can become much more complex. So let us investigate a problem where our r does not cancel out so nicely. We will find that sketching our domain is crucial!

Consider the following problem

Woha, this problem is introducing 2 equations on top of an integral!? Don't fret because while this may seem to be an overload of information, it actually isn't anything new.Evaluate: where S is the region bounded by the line and the circle in the first quadrant

First, we have to find our finite region. Since we are bounded by 2 equations we should draw these and figure out what the shape looks like.

is nothing more then a circle of radius a. So draw that on paper.

The line is simply a line that goes through the origin, that will eventually hit the circle. Notice how how the region bounded by our 2 equations (the circle and the line) is the entire area of the circle UP UNTIL the line? We don't want the area overtop of this line! We want the area underneath but contained within the circle.

But how are we going to compute this? Should we keep in cartesian coordinates or polar? Since we have a circle as part of our domain it might be best to transform to polar coordinates, so for this problem, we will use polar.

So lets find our domain for both theta and r.

Unfortunately, our theta domain is no longer ! Recall that for our earlier problems our theta domain was 360 degrees, but that's because we wanted to go the entire way around the circle!

In this case, we only want to go sweep from the x-axis to our line. Well, our line has an angle of ! We know this because of special triangles. If you notice the rise over run of one of the special triangles is so we can use that result: Special right triangles - Wikipedia, the free encyclopedia

We now know our theta bounds! in case you forget, our angle is measured from the x-axis in a counter clockwize manner.

But what about our bounds for r? Well...We go from the origin to the radius of the circle, so our r bounds must be from 0 to a.

Thus,

Note how we sub in and

So far we have dealt with static domains, and what I mean by that is r does not change with respect to theta. In the case of the above problem we are always drawing a line from the origin to the radius of the circle and sweeping that over the region.

But what if we have a shape where r is a function of theta? How does this change the problem?

Consider the following problem

If we draw the domain we can clearly see this is a triangle with the hypotenuse being bounded by the line this means our angle theta is by special triangles.Evaluate: where T is the region bounded by the triangle with the verticies (0,0), (1,0), (1,1)

So right away, we have our theta bounds

But what about r? If you think about it, the length of the hypotenuse will change with the inclination, so we need to find a value of r such that is relates to theta!

Draw another triangle and label the angle theta, the bottem of the triangle 1, and the hypotenuse r. Now take,

Which leads to

The reason why the bottem of the triangle is 1 is because this will never change, no matter what our x value is 1 and it's only the length of our hypotenuse that changes.

We can now compute this problem

Let and

Keeping with the idea of shifting our domain, what about a circle that isn't centered at the origin? Thus far, we have computed problems using polar co-ordinates where r is from the origin to the radius of the circle. Clearly, this won't work if the circle is centered somewhere else other then the origin!

Consider the following problem

First thing to note is that our cylinder is no longer centered at the origin.Find the volume bounded by the cylinder and the hemisphere

So we can see that our circle is centered on the x-axis at the coordinate

Not a problem, let's find our r domain

Remember that

So

So our r bounds are

Our integral becomes,

Where D is the domain of the circle in the first quadrant.

Take a moment to look at the picture. We want the volume between the hemisphere and the cylinder. Which is to say, we want the sphere projected over the domain of the cylinder.

I know I know, this isn't making a whole lot of sense but think of it like this,

If D is the domain of the hemisphere , this represents the volume of a hemisphere inside a hemisphere. However, if D represents a circle (or a cylinder) this integral represents the volume of a hemisphere INSIDE the domain of a cylinder.

The 4 in

comes from our bounds, we are going to bound ourselves to the first quadrant and multiply by 4 to get the entire hemisphere.

So to compute,

Note that our theta is bounded by 0 and because our value for r, that being is valid for the region

If you notice,

Let and

Let and

That's it for the instruction on basic polar coordinates dealing with double integrals. I will leave some homework problems for you!

1: Find the area between the paraboloids and

Spoiler:

2: Evaluate: where S is the area that lies outside the circle and the line

Spoiler:

3: Evaluate:

Spoiler: