Section 15.4 : Double Integrals in Polar Coordinates
1. Evaluate ∬Dy2+3xdA where D is the region in the 3rd quadrant between x2+y2=1 and x2+y2=9.
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Start SolutionBelow is a quick sketch of the region D.

For double integrals in which polar coordinates are going to be used the sketch of D is often not as useful as for a general region.
However, if nothing else, it does make it clear that polar coordinates will be needed for this problem. Describing this region in terms of Cartesian coordinates is possible but it would take two integrals to do the problem and the most of the limits will involve roots which often (not always, but often) leads to messy integral work.
The sketch shows that the region is at least partially circular and that should always indicate that polar coordinates are not a bad thing to at least think about. In this case, because of the Cartesian limits as discussed above polar coordinates are pretty much the only easy way to do this integral.
Note as well that once we have the sketch determining the polar limits should be pretty simple.
Show Step 2Okay, D is just a portion of a ring and so setting up the limits shouldn’t be too difficult. Here they are,
π≤θ≤32π1≤r≤3 Show Step 3The integral in terms of polar coordinates is then,
∬Dy2+3xdA=∫32ππ∫31((rsinθ)2+3rcosθ)rdrdθ=∫32ππ∫31r3sin2θ+3r2cosθdrdθWhen converting the integral don’t forget to convert the x and y into polar coordinates. Also, don’t forget that dA=rdrdθ and so we’ll pickup an extra r in the integrand. Forgetting the extra r is one of the most common mistakes with these kinds of problems.
Show Step 4Here is the r integration.
∬Dy2+3xdA=∫32ππ∫31r3sin2θ+3r2cosθdrdθ=∫32ππ(14r4sin2θ+r3cosθ)|31dθ=∫32ππ20sin2θ+26cosθdθ Show Step 5Finally, here is the θ integration.
∬Dy2+3xdA=∫32ππ20sin2θ+26cosθdθ=∫32ππ10(1−cos(2θ))+26cosθdθ=(10θ−5sin(2θ)+26sinθ)|32ππ=5π−26You’ll be seeing a fair amount of cos2θ, sin2θ and sinθcosθ terms in polar integrals so make sure that you know how to integrate these terms! In this case we used a half angle formula to reduce the sin2θ into something we could integrate.