 #1
happyparticle
 359
 19
 Homework Statement:
 Show that ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##
 Relevant Equations:

##d\vec{a} = R^2sin\theta d\theta d\phi \hat{r}##
##\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}##
##d = \vec{r}  \vec{R}## and ##\vec{r} = r \hat{z}##
Hey!
I have to show that the integral of the area of a sphere ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##, with ##d = \vec{r}  \vec{R}## and ##\vec{r} = r \hat{z}##
This is what I did.
##d\vec{a} = R^2sin\theta d\theta d\phi \hat{r}##, ##\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}##
Thus,
##\int \frac{d\vec{a}}{d} = \int_0^{2\pi} \int_0^{\pi} \frac{R^2 sin \theta (sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}) d\theta d\phi}{d}##
From here, I can separate this integral into 3 integrals. 2 of 3 = 0 and the last one = ## \frac{2R^2 \pi \hat{z}}{d}##
##= \frac{2R^2 \pi \hat{z}}{\vec{r}  \vec{R}}##
I don't see any error with my integrals. I don't know what I missed. I'm not sure if this is the right way to solve this kind of problem.
I have to show that the integral of the area of a sphere ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##, with ##d = \vec{r}  \vec{R}## and ##\vec{r} = r \hat{z}##
This is what I did.
##d\vec{a} = R^2sin\theta d\theta d\phi \hat{r}##, ##\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}##
Thus,
##\int \frac{d\vec{a}}{d} = \int_0^{2\pi} \int_0^{\pi} \frac{R^2 sin \theta (sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}) d\theta d\phi}{d}##
From here, I can separate this integral into 3 integrals. 2 of 3 = 0 and the last one = ## \frac{2R^2 \pi \hat{z}}{d}##
##= \frac{2R^2 \pi \hat{z}}{\vec{r}  \vec{R}}##
I don't see any error with my integrals. I don't know what I missed. I'm not sure if this is the right way to solve this kind of problem.