The boundary of the internal hole is given by $r = a + b\cos(4α)$ where $a > b >0$. The external boundary of the propeller is given by $r = c + d\cos(3α)$ where $c - d > a + b$ and $d > 0$.
1/Calculate the total area of the propeller as shown by the shaded region
2/calculate the average temperature of the propeller, given by $T (r, α) = C/r$, where $C$ is a constant.
3/Evaluate the average temperature numerically for a=3, b=1, c=12, d=5 and C=250 where distance is in cm and temperature is Celsius
I got an idea but don't know if it's right and when I ask my friend none of them know how to :( Are you suppose to integrate $r = c + d\cos(3α)$ then minus the hole $r = a + b\cos(4α)$ ?
$${1\over2}\int_0^{2\pi}\int_0^{p+q\cos(n\phi)} r^2\>dr\ d\phi\>,\qquad {1\over2}\int_0^{2\pi}\int_0^{p+q\cos(n\phi)}{C\over r}\> r^2\>dr\ d\phi\ .$$ since the area of the propeller in polar coordinate is: $${\rm d(area)}={1\over2} r^2{\rm d}(r,\phi)\ .$$
If someone help me solve it, it would make my day because I have no idea how to do it! :( I'll thank you if you could solve it! Pretty please.
Hints: Your strategy of subtracting areas is correct. However, it makes the calculation formally simpler to use $$ \int_{0}^{2\pi} \int_{r_{1}(\phi)}^{r_{2}(\phi)} r\, dr\, d\phi = \frac{1}{2} \int_{0}^{2\pi} \bigl(r_{2}(\phi)^{2} - r_{1}(\phi)^{2}\bigr)\, d\phi $$ instead of subtracting integrals. (You have an extra factor of $r$ in your integrands.)
Your set-up for the average temperature is on the right track (loosely, your integral represents "the total temperature" once you remove the extra factor of $r$), but you need to divide by the area to get the average temperature.