integrationmultivariable-calculus
I'm trying to calculate this integral but I'm a bit stuck. Has anyone got any tips/tricks to deal with the $e^{ir\cosθ}$ part?
$$\iiint r^{2}e^{ir\cos\theta}\sin\theta \,dr\,d\theta \,d\phi$$
Limits: $0\leq r \leq a$, $0\leq\theta\leq \pi$, $0\leq\phi\leq2\pi$.
I'm a first year chemistry student so keep the maths as simple as possible!
Your can use a simpler (?) transformation with cylindrical coordinates. Given that $x^2+(y-1)^2=1$ is a cylinder shifted 1 unit along the $y$ axis, you should not consider points in the $y<0$ plane ($\pi \le \theta\le 2\pi$), and the polar equation of the cylinder is not $r=1$ but $r=2\sin\theta$. Therefore, it should be $$ A=\{(r,\theta,z)\;|\; 0 \le \theta \le \pi, 0 \le r \le 2\sin\theta, 0\le z \le 2\} $$
Once you have done that, the integral is easy and equals: $$ \int_0^{\pi} \int_0^{2\sin\theta} \int_0^{2} z\, r\, r \; dz drd\theta = \frac{64}{9} $$
The required region is constituted of $3$ parts.
My notation: $\theta$ = Polar angle, where angle is taken with respect to the vector joining the two spheres, pointing two origin, $\phi$ = Azimuthal angle.
Parametrization of upper spherical cap: $$ \begin{pmatrix} &2 \sin(\theta) \cos(\phi) \\ & 2 \sin(\theta) \sin(\phi)\\ &2 \cos(\theta) \end{pmatrix}$$ with limits $ \theta \in \bigg ( \dfrac{\pi}{2} ,\dfrac{3 \pi}{2} \bigg) $,$ \phi \in (0, 2 \pi )$
Parametrization of lower spherical cap: $$ \begin{pmatrix} &G \sin(\theta) \cos(\phi) \\ & G \sin(\theta) \sin(\phi)\\ &G \cos(\theta) \end{pmatrix}$$ Where $$G = 2 \cos \theta - \sqrt{4 \cos^2 \theta -3}$$
with limits: $ \theta \in \bigg ( 0 , \arccos \bigg( \cfrac{ 2 - 2^{-1/2}}{\sqrt{5 - 2 \sqrt{2} } } \bigg) $,$ \phi \in (0, 2 \pi )$
Parametrization of conical section: $$ \begin{pmatrix} &H \sin(\theta) \cos(\phi) \\ & H \sin(\theta) \sin(\phi)\\ &H \cos(\theta) \end{pmatrix}$$
Where $$H = \cfrac{2}{\cos \theta + \sin \theta} $$ with limits: $ \theta \in \bigg ( \arccos \bigg( \cfrac{ 2 - 2^{-1/2}}{\sqrt{5 - 2 \sqrt{2} } }\bigg) , \dfrac{ \pi}{2} \bigg) $,$ \phi \in (0, 2 \pi )$
Best Answer
Your integral can be rewritten (by Fubini): $$ \left(\int_{\phi=0}^{2\pi}d\phi\right)\left(\int_{r=0}^ar^2\left(\int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta\right)dr\right) $$ Of course, the first factor is $2\pi$. Now for every $r>0$, do the change of variable $u=ir\cos\theta$, $du=-ir\sin\theta d\theta$ in the middle integral to get $$ \int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta=\int_{ir}^{-ir}e^u\frac{-du}{ir}=\frac{1}{ir}\int_{-ir}^{ir}e^udu=\frac{1}{ir} e^u\rvert_{-ir}^{ir}= \frac{1}{ir}(e^{ir}-e^{-ir}). $$ Now $$ \int_{r=0}^ar^2\left(\int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta\right)dr=\frac{1}{i}\int_0^ar(e^{ir}-e^{-ir})dr=2\int_0^ar\sin r dr $$ by Euler's formula $e^{ir}-e^{-ir}=2i\sin r$.
It only remains to integrate by parts $$ \int_0^ar\sin dr=(-r\cos r)\rvert_0^a+\int_0^a\cos r dr=-a\cos a+\sin r\rvert_0^a=-a\cos a+\sin a. $$ Finally, your integral is worth $$ 2\pi\cdot2(\sin a-a\cos a)=4\pi(\sin a -a\cos a). $$