Find the area that the curve encloses and then sketch it. r = 3 + 8 sin(6)
1 answer:
Answer:

Step-by-step explanation:
I assume you mean
:
Use the formula
where
and
are the lower and upper bounds and
is the equation of the polar curve.
Since the graph is symmetrical about the line
, let the bounds of integration be
to find half the area of the curve, and then find twice of that area:


![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](https://tex.z-dn.net/?f=A%3D%5Cbiggr%5B41%5Cbiggr%28%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29-48%5Ccos%5Cbiggr%28%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29-16%5Csin2%5Cbiggr%28%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D-%5Cbiggr%5B41%5Cbiggr%28-%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29-48%5Ccos%5Cbiggr%28-%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29-16%5Csin2%5Cbiggr%28-%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D%5C%5C%5C%5CA%3D%5Cbiggr%5B%5Cfrac%7B41%5Cpi%7D%7B2%7D-24%5Csqrt%7B2%7D%5Cbiggr%5D-%5Cbiggr%5B-%5Cfrac%7B41%5Cpi%7D%7B2%7D%2B24%5Csqrt%7B2%7D%5Cbiggr%5D%5C%5C%20%5C%5CA%3D41%5Cpi%5C%5C%5C%5CA%5Capprox128.8053)
Thus, the area of the curve is 41π square units. See below for a graph of the curve and its shaded area.
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