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Michael Corral: Vector Calculus

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3.5 Change of Variables in Multiple Integrals 121<br />

The change of variables formula can be used to evaluate double integrals in polar<br />

coordinates. Letting<br />

x= x(r,θ)=rcosθ and y=y(r,θ)=rsinθ,<br />

we have<br />

∂x ∂x<br />

J(u,v)=<br />

∂r ∂θ<br />

∂y ∂y<br />

=<br />

cosθ −rsinθ<br />

∣<br />

∣ sinθ rcosθ<br />

∂r ∂θ<br />

∣<br />

so we have the following formula:<br />

∣ = rcos2 θ+rsin 2 θ=r⇒|J(u,v)|=|r|=r,<br />

Double Integral in Polar Coordinates<br />

<br />

f(x,y)dxdy=<br />

R<br />

R ′ f(rcosθ,rsinθ)rdrdθ, (3.24)<br />

where the mapping x=rcosθ, y=rsinθ maps the region R ′ in the rθ-plane onto the<br />

region R in the xy-plane in a one-to-one manner.<br />

Example 3.10. Find the volume V inside the paraboloid z= x 2 +y 2 for 0≤z≤1.<br />

Solution: Using vertical slices, we see that<br />

<br />

V= (1−z)dA= (1−(x 2 +y 2 ))dA,<br />

R<br />

where R={(x,y) : x 2 + y 2 ≤ 1} is the unit disk in 2<br />

(seeFigure3.5.2). Inpolarcoordinates(r,θ)weknow<br />

that x 2 + y 2 = r 2 and that the unit disk R is the set<br />

R ′ ={(r,θ) : 0≤r≤1,0≤θ≤2π}. Thus,<br />

V=<br />

=<br />

=<br />

=<br />

∫ 2π ∫ 1<br />

0 0<br />

∫ 2π ∫ 1<br />

0<br />

∫ 2π<br />

0<br />

∫ 2π<br />

= π 2<br />

0<br />

0<br />

R<br />

(1−r 2 )rdrdθ<br />

(r−r 3 )drdθ<br />

(<br />

r 2 2 − r4 4<br />

1<br />

4 dθ<br />

∣<br />

∣ r=1<br />

r=0<br />

)<br />

dθ<br />

1<br />

z<br />

x 2 +y 2 = 1<br />

y<br />

0<br />

x<br />

Figure 3.5.2 z= x 2 +y 2

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