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

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

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

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)

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

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

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)

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

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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Math 200 Example

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

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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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

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

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

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