Speaker A
Welcome to an intro to double integrals in polar coordinates. Sometimes a double integral is much easier to evaluate using polar coordinates.
Learn how to convert and evaluate double integrals in polar coordinates with step-by-step examples and explanations.
Key Takeaways
- Polar coordinates simplify double integrals when the region is circular or defined by radius and angle.
- The differential area element in polar coordinates includes an extra factor of r, representing the arc length.
- Limits of integration must be carefully chosen based on the region's geometry in polar form.
- U-substitution is a useful technique for integrating functions involving r in polar coordinates.
- Understanding the geometric interpretation of polar integration aids in setting up and solving integrals correctly.
What the video covers
- Introduction to the advantages of using polar coordinates for double integrals, especially when the region is defined by polar equations.
- Explanation of converting rectangular coordinates (x, y) to polar coordinates (r, θ) using x = r cos θ and y = r sin θ.
- Derivation of the differential area element dA in polar coordinates as r dr dθ, including geometric reasoning using arc length.
- Step-by-step example calculating the volume under a surface over an annular region between two circles using polar coordinates.
- Setting up limits of integration for r and θ based on the region of integration in polar form.
- Performing integration with substitution techniques (u-substitution) for more complex integrals involving polar coordinates.
- Example of calculating volume over a quarter circle (first octant) region using polar coordinates and integral evaluation.
- Detailed algebraic simplifications and final volume results expressed in terms of π.
- Emphasis on the importance of the extra factor r in the integrand when converting to polar coordinates.
- Preview of additional examples to be covered in subsequent videos.
Chapters
- 00:00Introduction to double integrals in polar coordinates
- 00:24Converting rectangular to polar coordinates
- 00:55Function notation in polar form
- 01:26Geometric interpretation of differential area in rectangular coordinates
- 01:58Geometric interpretation of differential area in polar coordinates
- 03:01Origin of the extra factor r in polar integrals
- 03:30Example 1: Volume under surface between two circles
- 04:06Setting up the double integral for example 1
- 05:06Evaluating the integral for example 1
- 06:44Example 2: Volume over a quarter circle region
Full Transcript — Download SRT & Markdown
Speaker A
This is especially true if the region of integration can be easily defined using a polar equation.
Speaker A
So because we're going to be converting from rectangular to polar coordinates, it is important that we remember the following equations.
Speaker A
So you may want to copy these into your notes. If f of x is continuous on a polar region defined as given here, then we can rewrite a double integral in rectangular form into polar form as we see here.
Speaker A
And there's a couple important things to notice here. First, our function f of x y is going to be converted into a polar equation as we see here.
Speaker A
The way we do this is we replace x with r cosine theta, and y with r sine theta.
Speaker A
And sometimes the function is written like this to represent polar form, and sometimes it's just written as f of r theta.
Speaker A
And there's one more very important thing to notice, differential A is replaced with r dr d theta, so there's an extra factor of r in the integrand when we convert from rectangular to polar form.
Speaker A
And we'll take a look at why that is in just a minute. Let's take a look at these two graphs below.
Speaker A
Notice in rectangular form, we were determining the volume of these boxes where the bases were squares.
Speaker A
Well in polar form, we still want to determine the volume of these boxes, but now the bases aren't quite rectangular, so it's going to take a modification of the formula to determine differential A.
Speaker A
Let's take a look at that. So again in rectangular form, we determined that differential A was equal to dx times dy because this represented the area of a base, which was a square or a rectangle.
Speaker A
Now if we take a look at the base of one of those boxes in polar form, it's going to look more like this.
Speaker A
So delta theta would be the change in the angle, as we see here, this shorter radius would be r1, and the longer radius would be r2, so to determine the length of this piece here and this piece here,
Speaker A
we'd actually have to use the arc length formula, which states that arc length s is equal to r times theta.
Speaker A
So this length here would be r1 theta, and this should be r2 theta. And the length of this segment here would be delta r, which we can set equal to r2 minus r1.
Speaker A
Now of course as we let the number of boxes approach infinity, r1 is going to be approximately equal to r2, which we'll let equal r.
Speaker A
So from here we can say that delta A, or the change in the area, would be approximately equal to r times delta theta, again because these are all approximately equal to each other, times delta r.
Speaker A
So now you can see where this extra factor of r comes from. Go back up here and look at differential A is equal to r dr d theta, well here's the extra factor of r, and then again as the number of boxes approaches infinity,
Speaker A
delta theta is approximately equal to differential theta, and delta r is approximately equal to differential r.
Speaker A
So hopefully this rough outline helps you understand why when converting to polar form we do have this extra factor of r in the integrand.
Speaker A
Let's go and take a look at an example. We want to determine the volume of the solid below the surface given by f of x y above the xy plane over the region bounded by x squared plus y squared
Speaker A
equals one and x squared plus y squared equals four. Well the first thing we should recognize here is that these two are circles, one with radius one and another with radius two, as we see graphed here in red.
Speaker A
So our region of integration is the region between these two circles. This is a great example of when it'd be easier to evaluate this using polar coordinates.
Speaker A
And here you can see the solid, and I've traced the region of integration in red in the xy plane.
Speaker A
So let's see if we can set this up. The first thing we need to do is rewrite the function as a polar equation, let's first write this as four minus the quantity x squared plus y squared.
Speaker A
The reason that's helpful is remember that x squared plus y squared is equal to r squared, so we can rewrite this in polar form as four minus r squared.
Speaker A
Now see if we can set up our double integral. Our function is now four minus r squared, remember now we're going to have an extra factor of r, and then dr d theta.
Speaker A
Remember r stands for the radius, well this circle here would be r equals one, and this circle here would be r equals two.
Speaker A
So the limits of integration for r would just be from one to two, and then because we want one complete revolution around the circle, the limits of integration of theta will be from zero to two pi.
Speaker A
Let's go and distribute the r. So four r minus r cubed, dr d theta.
Speaker A
Let's go ahead and finish this on the next slide. So four times r squared over two minus r to the fourth over four, let's go ahead and clean this up a little bit, so we'll have two r squared minus
Speaker A
one-fourth r to the fourth. So when r is two it looks like we'll have eight minus, this is going to be 16 divided by four, that's four, and then when r is one we'll have two minus one-fourth.
Speaker A
So four minus one and three-fourths or seven-fourths, that's going to give us nine-fourths. Now [speaking indistinctly] theta, that's going to be nine-fourths theta, so when theta is two pi, we'll have two pi and then minus zero, give us nine halves pi.
Speaker A
So the volume of the solid over this region in orange would be nine halves pi.
Speaker A
Let's take a look at one more example. Here we want to determine the volume of this function over the region x squared plus y squared is less than or equal to four in the first octant.
Speaker A
So we have a circle with radius two, and we only want it in the first octant, so it's going to be this region here.
Speaker A
Now that we have a region of integration, first thing we should do is convert this to a polar equation.
Speaker A
So if we rewrite this as the square root of nine minus the quantity x squared plus y squared, we can replace x squared plus y squared with r squared.
Speaker A
So our double integral, the function is going to be the square root of nine minus r squared, and then don't forget our extra factor of r and then dr d theta.
Speaker A
So let's determine what r would be. Well this is a circle with radius two, so r is going to be from zero to two.
Speaker A
But now because we're in the first octant, theta is only going to be from zero to pi over two.
Speaker A
And now we're ready to go. This is going to require u substitution, though. [speaking indistinctly] have u equals nine minus r squared, so du's going to be negative two r dr, so that tells us that we can replace r dr with negative one half du.
Speaker A
So you're going to have negative one half du, and this'll just be u to the one-half.
Speaker A
So we'll have negative one half, see, u to the three-halves divided by three-halves, or times two-thirds, and u is really nine minus r squared, this simplifies here to negative one-third.
Speaker A
Let's finish this on the next slide. So when r is equal to two, we're going to have nine minus four, that's going to be five to the three-halves, and then when r is zero, we're going to have nine to the three-halves.
Speaker A
Let's go ahead and distribute here. Negative one-third times negative nine to the three-halves, that's going to give us a positive nine, and we'll have minus, and here we'd have five square root five, all over three.
Speaker A
This just the constant, just going to have nine minus five square root five over three, theta, times pi over two.
Speaker A
Let's go ahead and get a common denominator here. So denominator would be six, and we'd have 27 minus five square root of five times pi.
Speaker A
And we'll leave it like that. We'll take a look at some more examples in the next video.
Speaker A
Thank you for watching.
Topics:double integralspolar coordinatescalculusmultivariable calculusintegrationu-substitutionvolume calculationpolar area elementmath tutorialMathispower4u
