Page 57 - Calculus for the Clueless

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Find area of lemniscate r = 9 cos 2θ.
From Example 7 we have four-quadrant symmetry. So we will find the area in the first quadrant and multiply by
4.

Here's a little trick about finding the area of a curve with petals.

Example 10—

Find the area of r = 4 sin 6θ.

Just like the previous example, the graph as 12 petals, but all cosines are symmetric with respect to the x-axis.
cos 0 = 1. We find the smallest positive θ that makes cos 6θ = 0.6θ = π/2. So θ = π/12. From 0 to π/12
represents one-half petal. So we have 24 ½ petals to get the area. The integral is

Here's another trick with this problem.

This has 12 petals. We will find the area of one of them and multiply by 12. To find the area of one petal, we
find the first two values; sin 6θ = 0.6θ = 0 or θ = 0. 6θ = π. θ = π/6. So the integral is

Note

I know this last trick is one almost no one will use. But I've left it in because my original editor, David
Beckwith, was so great. This is one of his favorite tricks. Thanks, David.

Example 11—

Find the area of r = 4 cos 6θ.

We know we can slide the curve y = f(x) + a units to the right by replacing x by x - a. In the same way, we can
rotate r = f(θ) through a counterclockwise angle +α by replacing θ by θ -α. Thus, by rotating our curve by 15º =
π/12 radians, r = 4 cos 6(θ - π/12) = 4 cos (6θ - π/2) = 4 sin 6θ, which is exactly the curve in Example 10!!!!!!
Rotating a curve doesn't affect its area. So... the answer must be the same as in Example 10!!

Just like with x-y coordinates, we would like to find the area between two curves. In the case of x - y, it was the
top curve minus the bottom curve or the right curve minus the left curve. In polar coordinates, it is the outside
curve minus the inside one. The formula is

where θ 1 and θ 2 are the intersection of the two curves.
Example 12—