Parity and integration in spherical coordinates

[KostasV]

Hello people!
I have ended up to this integral $\int_{φ=0}^{2π} \int_{θ=0}^π \sin θ \ \cos θ~Y_{00}^*~Y_{00}~dθ \, dφ$ while I was solving a problem.
I know that in spherical coordinates when $\vec r → -\vec r$ :
1) The magnitude of $\vec r$ does not change : $r' → r$
2) The angles $θ$ and $φ$ change like $θ' → π-θ$ and $φ' → π+φ$
3) So parity of spherical harmonics is $\hat P$ $Y_{lm}(θ,φ)$ $=$ $Y_{lm}(θ',φ')=(-1)^l$ $Y_{lm}(θ,φ)$
4) Parity of $\cos θ$ and $\sin θ$ are $\cos θ'$ $=$ $(-1)$ $\cos θ$ and $\sin θ'$ $=$ $\sin θ$ respectively.
This means that in my case the integrated function has Parity equal to $(-1)$ .

So my question is:
Can I say that this integral is zero because of the odd (=parity is equal to (-1)) integrated funtion? Because in the xy plane when we integrate an odd function $F(x)=-F(-x)$ in a symmetric space (e.g. $\int_{-a}^a F(x) \, dx$ with F being odd) we can say that it is zero without to calculate it.
If yes, can I do this in general? meaning, if i get a random function in spherical coordinates which depends only from angles θ and φ and i want to integrate it with these limits: $\int_{φ=0}^{2π} \int_{θ=0}^π randomF(θ,φ) \, dθ \, dφ$ , can i find its parity and say if it is zero or not?

 

[vanhees71]

Well, $\mathrm{Y}_{00}=1/\sqrt{4 \pi}=\text{const}$ and your integral thus is
$$\propto \int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \cos \vartheta=\int_0^{\pi} \mathrm{d} \vartheta \frac{1}{2} \sin(2 \vartheta)= \left .-\frac{1}{4} \cos(2 \vartheta)\right|_0^{\pi}=0.$$

 

[KostasV]

Well, $\mathrm{Y}_{00}=1/\sqrt{4 \pi}=\text{const}$ and your integral thus is
$$\propto \int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \cos \vartheta=\int_0^{\pi} \mathrm{d} \vartheta \frac{1}{2} \sin(2 \vartheta)= \left .-\frac{1}{4} \cos(2 \vartheta)\right|_0^{\pi}=0.$$

Ok that was an easy integral but what if i have something more complicated like $\int_{φ=0}^{2π} \int_{θ=0}^π \sin θ \ \cos θ~Y_{11}^*~Y_{1-1}~dθ \, dφ$ ? If we use parity here we see that the parity of the integrated function is (-1) and the integral must be zero! No calculations ! Just used parity !
My question is not how to solve the integral making calculations . I want to solve it using parity and i want to tell me if my thoughts are correct on how to use parity in order to solve these integrals!
Thanks btw for your response :)

 

[blue_leaf77]

Can I say that this integral is zero because of the odd (=parity is equal to (-1)) integrated funtion?

Yes you can. Just remember that not every function has definite parity.

 

[KostasV]

Yes you can. Just remember that not every function has definite parity.

Thank you very much ! :)