What single transformation was applied to quadrilateral A to get quadrilateral B?

Find the area that the curve encloses and then sketch it.<br> r = 3 + 8 sin(6)

Rudiy27

Answer:

$A=41\pi\: \text{units}^2\approxA\approx128.8053\:\text{units}^2$

Step-by-step explanation:

I assume you mean $r=3+8\sin\theta$ :

Use the formula $\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta$ where $a$ and $b$ are the lower and upper bounds and $r(\theta)$ is the equation of the polar curve.

Since the graph is symmetrical about the line $\displaystyle \theta=\frac{\pi}{2}$ , let the bounds of integration be $\displaystyle \biggr(-\frac{\pi}{2},\frac{\pi}{2}\biggr)$ to find half the area of the curve, and then find twice of that area:

$\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta\\\\A=2\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2} {(3+8\sin\theta)^2} \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\sin^2\theta \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\biggr(\frac{1-\cos2\theta}{2} \biggr) \, d\theta\\\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} (9+48\sin\theta+32-32\cos2\theta) \, d\theta$

$\displaystyle A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} (41+48\sin\theta-32\cos2\theta) \, d\theta\\\\A=41\theta-48\cos\theta-16\sin2\theta\biggr|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\\\$

$A=\biggr[41\biggr(\frac{\pi}{2}\biggr)-48\cos\biggr(\frac{\pi}{2}\biggr)-16\sin2\biggr(\frac{\pi}{2}\biggr)\biggr]-\biggr[41\biggr(-\frac{\pi}{2}\biggr)-48\cos\biggr(-\frac{\pi}{2}\biggr)-16\sin2\biggr(-\frac{\pi}{2}\biggr)\biggr]\\\\A=\biggr[\frac{41\pi}{2}-24\sqrt{2}\biggr]-\biggr[-\frac{41\pi}{2}+24\sqrt{2}\biggr]\\ \\A=41\pi\\\\A\approx128.8053$