# Double Integrals in Polar Coordinates – Example 2

Welcome to a second example of converting a double integral in rectangular form to polar form. As we discussed in a previous video. To convert a double integral in
rectangular form to polar form. We have to convert the function f of x, y into a function in terms of r and theta. And differential A is replaced with an extra factor of r, dr, d theta. So we can’t forget about this extra factor of r when converting to polar form. Let’s go and take a look at our second example. Let’s see if we can determine the region of integration. We’re integrating first with respect to y which means y has to be greater than or equal to zero and less than or equal to the
square root of x minus x squared. So y has to be greater than or equal to zero. And less or equal to the square root of x minus x squared. And the outer integral we’re integrating with respect to x which means
x has to be greater than or equal to zero and less than or equal to one. Let’s see if we can determined
the region of integration from this. Well we know x is positive and y is positive
so we’re dealing with the first quadrant. So let’s focus on determining the graph of y is less than or equal to the square root of x minus x squared. One thing we could do is square both sides. That would give us y squared is less than or equal to x minus x squared. If we add x squared to both
sides we have x squared plus y squared less than or equal to x. So we might wanna type this into our graphing
calculator but we could also try to do this by hand. By making a T table. So were gonna go ahead and see if
we can graph x squared plus y squared equals x. By making a T table. Well if x is zero. Course we have zero on both
sides so y would have to be zero. And if x was one. We have a one on both sides y would have to be zero. Let’s focus on what would happen if x was one half. Well if x is one half we have the equation one fourth plus y squared must equal one half. So if we subtract one fourth on both
sides we have y squared equals one fourth. And then y would have to be equal to one half. So from this you might be able to
determine that we have to graph that looks like this. Here’s the point zero, zero. Here’s the point one half, one half. Here’s the point one, zero. So because y has to be less than this semi circle. We would shade below meaning this would be our region of integration. Let’s see if we can start to set
up our double integral from here. I’m gonna go ahead and label this curve it was a curve y equals the square root of x minus x squared. So now we have our function f of x, y equals x times y. Let’s see if we can determine f of r, theta. Meaning convert this function into polar form. Member x is equal to r cosine theta. And y is equal to r sine theta. So f of r, theta. Would be r squared cosine theta sine theta. So we can go to our double integral in polar form. Here’s f of r, theta. So we have r squared. Cosine theta, sine theta. And don’t forget we have an extra factor of r and then dr, d theta. So now we need to determine the limits of integration. First respect to r. And then with respect to theta. And so this one it’s gonna be a little bit tricky
there’s a couple ways are going about doing it. If we don’t recognize the polar equation that we create. This semi circle here. You go ahead and take this curve here. And rewrite it as a polar equation. So let’s go ahead and do that we would have y equals the square root of x minus x squared. And y is equal to r sine theta. We have the square root again x is r cosine theta. And we have minus r squared. Cosine squared theta. Now if we squared both sides of this
equation we’re gonna have r squared sine squared theta. Would equal r cosine theta. Minus r squared cosine squared theta. Let’s go and add r squared cosine squared to both sides that will give us r squared sine squared theta. Plus r squared cosine squared theta. Equals r cosine theta. Now if we factor out r squared we’re gonna have sine squared plus cosine
squared but that’s equal to one. And divide both sides by r. You have r equals cosine theta. So this tells us that the radius. This tells us we could use this polar
equation to trace out the radius. So the limits of integration with
respect to r would be from zero to cosine theta. Now let’s go ahead and verify
this with our graphing calculator. To write down as I put the calculator into green mode because it is easier to read and also polar form. And I typed in r equals cosine theta. Let’s take a look at the graph. Notice it does generate this half circle as
well as the second half in the fourth quadrant. Let’s see what theta would be. If we press trace. We can see that when theta is equal to zero. So we be at this point here. And as we increase theta. You can see it does trace out
the semi circle that we want. And when theta is equal to 90 degrees or pi over two. We’re back at the origin. So that works perfectly. That verifies our limits of integration for r. It also tells us that theta would
be from zero to pi over two. And notice for the limits of integration for r. There was in terms of function
that had to be in terms of theta. Let’s go and finish this on the next slide. Notice we are going to have r to the third
power because of this extra factor of r. And now we’re gonna integrate with
respect to r treating theta as a constant. So we have r to the fourth over four. Go ahead and pull out the one fourth. Now we want to replace r with cosine theta. It’s gonna give us cosine to the fourth theta here plus another factor of cosine theta we’ll have cosine to the fifth theta sine theta. Then when r is zero this should be zero. Now we’re gonna have to perform u
substitution if we let u equal cosine theta. And differential u is gonna be
equal to negative sine theta d theta. So we’re gonna replace sine
theta, d theta with negative du. So negative du takes the place of sine theta, d theta. This would be u to the fifth. So we have negative one fourth
times u to the six over six. Let’s go rewrite this in terms of theta
we have negative one twenty fourth. This should be cosine to the sixth theta. Well cosine pi over two would be zero. So we have zero to the sixth is zero. Minus cosine to the zero which is one. One to the sixth is one. So we have negative one twenty fourth times a negative one or positive one twenty fourth. That will do it for this second example. I hope you found this helpful.

## 27 thoughts on “Double Integrals in Polar Coordinates – Example 2”

1. piemaster4 says:

from 4:08-5:13 I just used the previous formula you had x^2+y^2=x to get r^2=rcostheta. then it works the same π

2. Jessica B. says:

i wish you were my teacher. <3 you

3. Ho samson says:

Thank You Very Much!

4. Jordan says:

You are amazing..

5. Rachel Goettling says:

I keep getting burned by these types of problems where the lower bound of integration for the internal integral is the line y = x. How do I convert this equation and solve for r? The only answer I come up with is r = r*tan(theta), which just turns out to be garbage since you're now trying to evaluate the partial integral by replacing an r with another r.

6. Devan Anderson says:

@rmwhite13691 Same problem here

7. 12Nowhere says:

at 6:04, '' and when theta is 90 degrees or pie/2 we are back at the origin'' ???i
when theta is 180 degrees or pie we are back at the origin.
whe theta is 90 degrees we are at ( 1/2,1/2).

8. NeverLieToYa says:

the point is the sketch is in the first quadrant which is 90 degrees..so it'll back at its origin π

9. Safaa Az says:

Thanks for your videos ,very helpful as usual and btw i like the quotes at the end, really inspiring. God bless you.

10. Abdul Azim says:

Many thanks mate… you are surely preparing me for my exam with your thorough steps π

11. Steeve Cantave says:

Well, y = x passes through the angle pi/4. if it was a circle centered at the origin r would go from 0 to whatever the radius is.

12. Fawzy Hegab says:

great video , I wonder what is the name of the programe to get this calculator you have used through the video ?

13. Daler Asrorov says:

Thanks a lot. Keep the videos going!

14. Terek Li says:

why isn't the limit of r not 0 to 1/2 considering that 1/2 is the radius of the region?

15. Sara A says:

@Mathispower4u Thank you for this video- just wondering why theta is equal to 90 or pi/2 and not pi? When I graph it in rectangular just like the previous example we were able to get theta from the graph. Thank you

16. Jazmin Sanchez says:

17. Noel Martinez says:

this guy is awesome. he goes at the perfect pace. some videos out there are way too slow.

18. CocoGras says:

Maybe I'm missing something, but 1/24 seems kind of small for the area of half of a circle with radius of 1/2. If area of a circle is pi*radius <squared, then wouldn't the area of this particular half circle be ~3/8?

19. Julio C. S says:

Your videos are saving my Calc III class. I will definitely will recommend it to my friends!

20. Eddy says:

this is better than patrickJMT and khan

21. Kushagra Sharma says:

U just saved my grades there mate….thnxπ

22. Nidhi Solanki says:

Why does the limit of theta vary from 0 to Β₯/2? Shouldn't it be from 0 to Β₯? (Β₯=pi)

23. John Valgon says:

why we use r=0 to r=cos(theta) instead of r=one half ? it is just a semicircle

24. Pranjal Arya says:

Nice explanation

25. Sabrina Camargo says:

I found it easier to replace the function that's being integrated with the square root of r^2 rather than plug in all that.
r^2 = x^2 + y^2

26. Nicacio Freire says: