Question

1. Consider the following polar curve: r = 3 + 2 cos θ (a) Sketch the curve. (b) Find the area it encloses. (c) Set up an integral that represents the length of one loop of the curve. 2. Consider the following polar curve: r = 4 cos 3θ (a) Sketch the curve. (b) Find the area it encloses. (c) Set up an integral that represents the length of one loop of the curve.

Answer 1

SEE THE PROCEDURE PLEASE

Explanation:

1.

a. The plot is attached below

b. The area is given by

`A=(1)/(2)\int_\alpha^\beta [r(\theta)]^2d\theta\\\\A=(1)/(2)\int_(0)^(2\pi)[3+2cos\theta]^2d\theta\\\\A=(1)/(2)\int_(0)^(2\pi)[9+12cos\theta+4cos^2\theta]d\theta\\\\A=(1)/(2)[9(2\pi)+12sin(2\pi)+2(2\pi)+sin(4\pi)]\\\\A=11\pi`

c.

`L=\int_0^(2\pi)\sqrt{r^2+((dr)/(d\theta))^2}d\theta\\\\L=\int_0^(2\pi)√((3+2cos\theta)^2+(-2sin\theta)^2)d\theta=\int_0^(2\pi)√(13+12cos\theta)d\theta`

2.

a. The plot is attached below

b. by symmetry:

`A=6*(1)/(2)\int_{-(\pi)/(6)}^{(pi)/(6)}16cos^2\theta d\theta\\\\A=6[4√(3)+(8\pi)/(3)]=24√(3)+16\pi`

c.

`L=6*\int_{-(\pi)/(6)}^{(\pi)/(6)}√(16cos^2\theta-144sin^23\theta) d\theta`

HOPE THIS HELPS!!