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Calculus - Multivariate≻Integration≻Average Value≻Polar

Table 6.4.7(a)   In polar coordinates, computation of average value by task template

Initialize

$F\=4-3 r \cos \left(\mathrm{\theta }\right)\+5 r \sin \left(\mathrm{\theta }\right)$$\stackrel{\text{assign}}{\to }$

Calculate the average value of $F$ 

$\frac{{\int }_{-\mathrm{\π}\/2}^{\mathrm{\π}\/2}{\int }_1^{1\+ \cos \left( \mathrm{\θ}\right)}F\cdot r \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}}{{\int }_{-\mathrm{\π}\/2}^{\mathrm{\π}\/2}{\int }_1^{1\+ \cos \left( \mathrm{\θ}\right)}r \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}}$ = $\frac{4-\frac{7}{8}\mathrm{\π}}{\frac{1}{4}\mathrm{\π}\+2}$$\stackrel{\text{normalize}}{\to }$$-\frac{1}{2}\frac{-32\+7\mathrm{\π}}{\mathrm{\π}\+8}$$\stackrel{\text{at 10 digits}}{\to }$$0.4491660992$

Table 6.4.7(b)   From first principles and in polar coordinates, calculation of average value

Initialize

$F≔4-3 r \cos \left(\mathrm{\theta }\right)\+5 r \sin \left(\mathrm{\theta }\right)\:$

Compute the average value of $F$ 

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(F\,r\=1..1\+\cos \left(\mathrm{\θ}\right)\,\mathrm{\θ}\=-\mathrm{\π}\/2..\mathrm{\π}\/2\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\,\mathrm{output}\=\mathrm{integral}\right)$

$\frac{{\∫}_{-\frac{1}{2}\mathrm{\π}}^{\frac{1}{2}\mathrm{\π}}{\∫}_1^{1\+\cos \left(\mathrm{\θ}\right)}\left(4-3r\cos \left(\mathrm{\θ}\right)\+5r\sin \left(\mathrm{\θ}\right)\right)r\ⅆr\ⅆ\mathrm{\θ}}{{\∫}_{-\frac{1}{2}\mathrm{\π}}^{\frac{1}{2}\mathrm{\π}}{\∫}_1^{1\+\cos \left(\mathrm{\θ}\right)}r\ⅆr\ⅆ\mathrm{\θ}}$

$q≔\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(F\,r\=1..1\+\cos \left(\mathrm{\theta }\right)\,\mathrm{\θ}\=-\mathrm{\π}\/2..\mathrm{\π}\/2\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\theta }\right]\right)$

$\frac{4-\frac{7}{8}\mathrm{\π}}{\frac{1}{4}\mathrm{\π}\+2}$

$\mathrm{normal}\left(q\right)$ = $-\frac{1}{2}\frac{-32\+7\mathrm{\π}}{\mathrm{\π}\+8}$$$

$\mathrm{evalf}\left(q\right)$ = $0.4491660992$$$