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Calculus - Multivariate≻Integration≻Average Value≻Polar

Table 6.4.9(a)   In polar coordinates, computation of average value by task template

Initialize

$F\=r^2 \left(2 {\cos }^2\left(\mathrm{\theta }\right)\+3 {\sin }^2\left(\mathrm{\theta }\right)\right)$$\stackrel{\text{assign}}{\to }$

Calculate the average value of $F$ 

$\frac{{\int }_{-\mathrm{\π}\/3}^{\mathrm{\π}\/3}{\int }_{1\+\cos \left(\mathrm{\θ}\right)}^{3 \cos \left( \mathrm{\θ}\right)}F\cdot r \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}}{{\int }_{-\mathrm{\π}\/3}^{\mathrm{\π}\/3}{\int }_{1\+\cos \left(\mathrm{\θ}\right)}^{3 \cos \left( \mathrm{\θ}\right)}r \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}}$ = $\frac{\frac{1521}{320}\sqrt{3}\+\frac{223}{24}\mathrm{\π}}{\mathrm{\π}}$$\stackrel{\text{expand}}{\=}$$\frac{1521}{320}\frac{\sqrt{3}}{\mathrm{\π}}\+\frac{223}{24}$$\stackrel{\text{at 10 digits}}{\to }$$11.91220182$

Table 6.4.9(b)   From first principles and in polar coordinates, calculation of average value

Initialize

$F≔r^2 \left(2 {\cos }^2\left(\mathrm{\theta }\right)\+3 {\sin }^2\left(\mathrm{\theta }\right)\right)\:$

Compute the average value of $F$ 

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(F\,r\=1\+\cos \left(\mathrm{\θ}\right)..3 \cos \left(\mathrm{\θ}\right)\,\mathrm{\θ}\=-\mathrm{\π}\/3..\mathrm{\π}\/3\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\θ}\right]\,\mathrm{output}\=\mathrm{integral}\right)$

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

$q≔\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(F\,r\=1\+\cos \left(\mathrm{\theta }\right)..3 \cos \left(\mathrm{\theta }\right)\,\mathrm{\θ}\=-\mathrm{\π}\/3..\mathrm{\π}\/3\,\mathrm{coordinates}\=\mathrm{polar}\left[r\,\mathrm{\theta }\right]\right)$

$\frac{\frac{1521}{320}\sqrt{3}\+\frac{223}{24}\mathrm{\π}}{\mathrm{\π}}$

$\mathrm{expand}\left(q\right)$ = $\frac{1521}{320}\frac{\sqrt{3}}{\mathrm{\π}}\+\frac{223}{24}$$$

$\mathrm{evalf}\left(q\right)$ = $11.91220182$$$