A curve is given in polar coordinates by
a) Sketch the curve.
b) Find the area enclosed by the curve.
| Foundations:
|
| Area under a polar curve
|
Solution:
(a)
(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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin(2\theta )}
, we have
|
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}
|
Return to Sample Exam