1. Double Integrals with Polar Coordinates
Math 200
Week 8 - Wednesday
Double Integrals with Polar Coordinates

2. Math 200
Goals
Be able to compute double integral calculations over non rectangular regions using polar coordinates
Be able to convert back and forth between polar and rectangular double integrals
Know when to use polar coordinates for double integrals

3. Review We first encountered polar coordinates in Calc II
Math 200
Review
We first encountered polar coordinates in Calc II
The polar axis matches up with the positive x-axis
We measure θ counterclockwise off of the polar axis
We measure r to be the distance from the origin in the direction of θ (r can be negative)
(x,y) ~ (r,θ) y x r θ

4. Double integrals with polar coordinates
Math 200
Double integrals with polar coordinates
In the previous section, we figured out how to find the volume bounded between a region in the plane and a surface.
We now want to set up double integrals in polar coordinates.
This requires breaking R up in polar coordinates.
Region: R
Surface: f(x,y)

5. We want to “chop R up” in terms of r and θ (rather than x and y)
Math 200
Consider the region R
We want to “chop R up” in terms of r and θ (rather than x and y)
Q: What is the area of each polar rectangle in terms of r and θ?
A: dA = rdrdθ
The extra r term comes from the arc length formula for a circle: L=rθ
rdθ
dA =rdrdθ dr

6. The setup In rectangular coordinates we replace dA with dxdy or dydx
Math 200
The setup
In rectangular coordinates we replace dA with dxdy or dydx
In polar coordinates, we have an extra term in dA
dA = rdrdθ or dA = rdθdr
Most often, it’ll be rdrdθ
For a polar region R bounded by two polar curves r1(θ) and r2(θ) and between θ1=α and θ2=β, we have

7. Example Let’s use polar coordinates to evaluate the integral
Math 200
Example
Let’s use polar coordinates to evaluate the integral
From the bounds we see that…
The region R is bounded below by y1(x) = 0 and bounded above by y2(x) = (16-x2)1/2
R extends from x1 = 0 to x2 = 4
y2(x)
y1(x)

8. How do we describe this same region in terms of r and θ?
Math 200
How do we describe this same region in terms of r and θ?
Draw a ray from the origin through the region
We enter the region right away (at the origin):
r1(θ) = 0
We exit the region through the circle of radius 4:
r2(θ) = 4
θ goes from 0 to π/2
r2(θ) = 4
r1(θ) = 0

9. We can’t forget the extra r term in dA
Math 200
We also need to rewrite the function (x2 + y2)1/2 in terms of r and θ
Recall that r2 = x2+y2
So, (x2 + y2)1/2 = (r2)1/2 = r
Putting it all together we have
We can’t forget the extra r term in dA

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12. Math 200
Example

13. y=x corresponds to θ=π/4 and θ=5π/4 So, θ1=π/4 and θ2=5π/4
Math 200
Let’s look at R first:
x2+y2≥1
x2+y2≤4
y≥x
Putting it all together
Rewriting this in terms of r isn’t so bad:
1≤x2+y2≤4
1≤r2≤4
1≤r≤2
So, r1 = 1 and r2 = 2
y=x corresponds to θ=π/4 and θ=5π/4
So, θ1=π/4 and θ2=5π/4

14. Putting it all together…
Math 200
Putting it all together…
We need to do u-substitution: Let u=-r2; then du = -2rdr
We can put the -1/2 out in front:

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16. Math 200
Example
Compute the volume of the solid bounded above by the paraboloid z = 9 - x2 - y2, bounded below by z = 0, and contained within the cylinder x2 - 3x + y2 = 0

17. The surface is f(x,y)=9-x2-y2
Math 200
In order to use a double integral to compute this volume, we need to think of the solid as bounded between a surface f(x,y) and a region in the xy- plane
The surface is f(x,y)=9-x2-y2
The region is the circle given by x2-3x+y2=0
Now we just have to interpret all this in polar coordinates

18. We should recognize this from polar coordinates in Calc II:
Math 200
Recall: r2 = x2 + y2
So, f(x,y) = 9 - x2 - y2 becomes z = 9 - r2
We also have that x = rcosθ
So, the region R: x2-3x+y2=0 becomes r2-3rcosθ = 0
We can rewrite this as r = 3cosθ
r1(θ) = 0
r2(θ) = 3cosθ
We should recognize this from polar coordinates in Calc II:
To fill in the circle, we need r to go from 0 to 3cosθ and we need θ to go from 0 to π

19. Math 200