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

Answer:

SEE THE PROCEDURE PLEASE

Step-by-step explanation:

1.

a. The plot is attached below

b. The area is given by

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

c.

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

2.

a. The plot is attached below

b. by symmetry:

$A=6*\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{pi}{6}}16cos^2\theta d\theta\\\\A=6[4\sqrt{3}+\frac{8\pi}{3}]=24\sqrt{3}+16\pi$

c.

$L=6*\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\sqrt{16cos^2\theta-144sin^23\theta} d\theta$

HOPE THIS HELPS!!