- #1
oliverkahn
- 27
- 2
Let:
\begin{align}
r&=\sqrt{a^2 + p^2 - 2ap \cos \theta}\\
s&=a\\
t&=p\\
f(r) &= \text{continuous function of } r\\
g(s) &= \text{continuous function of } s\\
\end{align}
Consider the expression:
\begin{align}
\int_{q'}^q \int_{b'}^b g(s)\ \int_{s-t}^{s+t} f(r)\ dr\ ds\ dt\
\end{align}
We next have to change the variables from ##(r,s,t)## to ##(\theta, a, p)##. The Jacobian of the coordinate transformation (after computing) is:
##J= \dfrac{\partial r}{\partial \theta}=\dfrac{a\ p\ \sin\theta}{r}##
Thus our new function becomes ##J\ f(r) =\dfrac{a\ p\ \sin\theta}{r} f(r)##
Question:
One of my friends said that the limits of the integration would be as follows:
\begin{align}
\int_{q'}^q \int_{b'}^b g(a)\ \int_{0}^{\pi} \dfrac{a\ p\ \sin\theta}{r} f(r) \ d \theta\ da\ dp\
\end{align}
Is he correct?
\begin{align}
r&=\sqrt{a^2 + p^2 - 2ap \cos \theta}\\
s&=a\\
t&=p\\
f(r) &= \text{continuous function of } r\\
g(s) &= \text{continuous function of } s\\
\end{align}
Consider the expression:
\begin{align}
\int_{q'}^q \int_{b'}^b g(s)\ \int_{s-t}^{s+t} f(r)\ dr\ ds\ dt\
\end{align}
We next have to change the variables from ##(r,s,t)## to ##(\theta, a, p)##. The Jacobian of the coordinate transformation (after computing) is:
##J= \dfrac{\partial r}{\partial \theta}=\dfrac{a\ p\ \sin\theta}{r}##
Thus our new function becomes ##J\ f(r) =\dfrac{a\ p\ \sin\theta}{r} f(r)##
Question:
One of my friends said that the limits of the integration would be as follows:
\begin{align}
\int_{q'}^q \int_{b'}^b g(a)\ \int_{0}^{\pi} \dfrac{a\ p\ \sin\theta}{r} f(r) \ d \theta\ da\ dp\
\end{align}
Is he correct?