- #1

happyparticle

- 405

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- 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.