009C Sample Final 1, Problem 8

A curve is given in polar coordinates by

$r=1+\sin 2\theta$

$0\leq \theta \leq 2\pi$

a) Sketch the curve.

b) Find the area enclosed by the curve.

Foundations:
Area under a polar curve

Solution:

(a)

Step 1:
Insert sketch

(b)

Step 1:
Since the graph has symmetry (as seen in the graph), the area of the curve is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 2\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {1}{2}}(1+\sin(2\theta )^{2})~d\theta }

Step 2:
Using the double angle formula for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(2\theta)} , we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2~d\theta} & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\sin^2(2\theta)~d\theta} \\ &&\\ & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\frac{1-\cos(4\theta)}{2}~d\theta}\\ &&\\ & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{3}{2}+2\sin(2\theta)-\frac{\cos(4\theta)}{2}~d\theta}\\ &&\\ & = & \displaystyle{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}}\\ \end{array}}

Step 3:
Lastly, we evaluate to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}} & = &\displaystyle{\frac{3}{2}\frac{3\pi}{4}-\cos\bigg(\frac{3\pi}{2}\bigg)-\frac{\sin(3\pi)}{8}-\bigg[\frac{3}{2}\bigg(-\frac{\pi}{4}\bigg)-\cos\bigg(-\frac{\pi}{2}\bigg)-\frac{\sin(-\pi)}{8}\bigg]}\\ &&\\ & = & \displaystyle{\frac{9\pi}{8}+\frac{3\pi}{8}}\\ &&\\ & = & \displaystyle{\frac{3\pi}{2}}\\ \end{array}}

Final Answer:
(a) See Step 1 above.
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3\pi}{2}}

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