Double Integrals with Polar Coordinates

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

2. Math 200 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 Goals

3. 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) y θ Math 200 Review x (x,y) ~ (r,θ) r

4. 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. Surface: f(x,y) Math 200 Double integrals with polar coordinates Region: R

5. 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 Math 200 rdθ dA =rdrdθ dr

6. Math 200 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 The setup

7. Math 200 Let’s use polar coordinates to evaluate the integral Example • 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. 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 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 Math 200

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13. 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. 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 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 Example

17. 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. 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 r2(θ) = 3cosθ r1(θ) = 0 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

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