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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r=f(\theta )} is given by
$\int _{\alpha _{1}}^{\alpha _{2}}{\frac {1}{2}}r^{2}~d\theta$ for appropriate values of $\alpha _{1},\alpha _{2}.$

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 (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 $\sin(2\theta ),$ 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
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}\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) 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}}

Return to Sample Exam