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:
The area under a polar curve $r=f(\theta )$ is given by
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 \int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta} for appropriate values of 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 \alpha_1,\alpha_2.}

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 $\sin(2\theta ),$ we have
${\begin{array}{rcl}\displaystyle {2\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {1}{2}}(1+\sin(2\theta ))^{2}~d\theta }&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}1+2\sin(2\theta )+\sin ^{2}(2\theta )~d\theta }\\&&\\&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}1+2\sin(2\theta )+{\frac {1-\cos(4\theta )}{2}}~d\theta }\\&&\\&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {3}{2}}+2\sin(2\theta )-{\frac {\cos(4\theta )}{2}}~d\theta }\\&&\\&=&\displaystyle {{\frac {3}{2}}\theta -\cos(2\theta )-{\frac {\sin(4\theta )}{8}}{\bigg |}_{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}.}\\\end{array}}$

Step 3:
Lastly, we evaluate to get
${\begin{array}{rcl}\displaystyle {{\frac {3}{2}}\theta -\cos(2\theta )-{\frac {\sin(4\theta )}{8}}{\bigg |}_{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}}&=&\displaystyle {{\frac {3}{2}}{\bigg (}{\frac {3\pi }{4}}{\bigg )}-\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}-{\frac {\sin(3\pi )}{8}}-{\bigg [}{\frac {3}{2}}{\bigg (}-{\frac {\pi }{4}}{\bigg )}-\cos {\bigg (}-{\frac {\pi }{2}}{\bigg )}-{\frac {\sin(-\pi )}{8}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {9\pi }{8}}+{\frac {3\pi }{8}}}\\&&\\&=&\displaystyle {{\frac {3\pi }{2}}.}\\\end{array}}$

Final Answer:
(a)     See Step 1 above.
(b)     ${\frac {3\pi }{2}}$

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