536 CHAPTER 27. THE INTEGRAL IN OTHER COORDINATES
This is easy to do in polar coordinates. The disk involved has θ going from 0 to 2π andr from 0 to 2. Therefore, the integral to work is just
∫ 2π
0
∫ a
0r
dA︷ ︸︸ ︷rdrdθ =
23
πa3
Notice how in these examples the circular disk is really a rectangle [0,2π]× [0,a]. This iswhy polar coordinates are so useful. The next example was worked earlier from a differentpoint of view.
Example 27.1.3 Find the area of the inside of the cardioid r = 1+ cosθ , θ ∈ [0,2π].
How would you go about setting this up in rectangular coordinates? It would be veryhard if not impossible, but is easy in polar coordinates. This is because in polar coordinatesthe region integrated over is the region below the curve in the following picture. It is oneof those regions which is simple to integrate over. The graph of the top is r = 1+ cosθ .However, the graph of the cardioid in rectangular coordinates is not at all simple. Seethe material on polar coordinates where graphs of cardioids were provided in rectangularcoordinates.
0 π
4 π3π
4 2π
The integral is ∫ 2π
0
∫ 1+cos(θ)
0rdrdθ =
32
π
Example 27.1.4 Let R denote the inside of the cardioid r = 1+ cosθ for θ ∈ [0,2π]. Find∫R
xdA
Here the convenient increment of area is rdrdθ and so the integral is∫ 2π
0
∫ 1+cos(θ)
0xrdrdθ
Now you need to change x to the right coordinates. Thus the integral equals∫ 2π
0
∫ 1+cos(θ)
0(r cos(θ))rdrdθ =
54
π
A case where this sort of problem occurs is when you find the mass of a plate given thedensity.
Definition 27.1.5 Suppose a material occupies a region of the plane R. The densityλ is a nonnegative function of position with the property that if B ⊆ R, then the mass of Bis given by
∫B λdA. In particular, this is true of B = R.
Example 27.1.6 Let R denote the inside of the polar curve r = 2+ sinθ . Let λ = 3+ x.Find the total mass of R.
As above, this is ∫ 2π
0
∫ 2+sin(θ)
0(3+ r cos(θ))rdrdθ =
272
π