Question

Use a double integral to find the area of the region that has the indicated shape. One loop of r = 2cos(4theta)

Answer 1

we need to find a smallest interval where cos 4t goes form 0 to 0

Answer 2

the other integral is just measureing r from 0 to the function itself

Answer 3

\[\text{when t=pi/2 and 3pi/2; cos t = 0}\]
\[\frac{4}{4}\frac{pi}{2}=4\frac{pi}{8}\]
\[\frac{4}{4}\frac{3pi}{2}=4\frac{3pi}{8}\]

so the smallest interval i see is from pi/8 to 3pi/8

Answer 4

\[\large \int_{t=pi/8}^{t=3pi/8}\int_{r=0}^{r=2cos(4t)}r\ dr.dt\]