009C Sample Final 1, Problem 8

A curve is given in polar coordinates by

$r=1+\sin 2\theta$

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 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 (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 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
${\begin{array}{rcl}\displaystyle {{\frac {3}{2}}\theta -\cos(2\theta )-{\frac {\sin(4\theta )}{8}}{\bigg |}_{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}}&=&\displaystyle {{\frac {3}{2}}{\frac {3\pi }{4}}-\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}-{\frac {\sin(3\pi )}{8}}-{\bigg [}{\frac {3}{2}}{\bigg (}-{\frac {\pi }{4}}{\bigg )}-\cos {\bigg (}-{\frac {\pi }{2}}{\bigg )}-{\frac {\sin(-\pi )}{8}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {9\pi }{8}}+{\frac {3\pi }{8}}}\\&&\\&=&\displaystyle {\frac {3\pi }{2}}\\\end{array}}$

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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