A Surface Integral

Calculus Level pending

Consider a sphere of radius R = 5 R=5 centered at the origin, and a point r 0 = ( 0 , 1 , 2 ) \mathbf{r_0} = (0, 1, 2) inside the sphere.

Calculate the surface integral

1 4 π ∫ S ( r − r 0 ) ⋅ d S \dfrac{1}{4 \pi} \displaystyle \int_S (\mathbf{r} - \mathbf{r_0}) \cdot d\mathbf{S}

Here r \mathbf{r} is the position vector of a point on the surface of the sphere.

25 64 98 125

Hosam Hajjir by Hosam Hajjir , 56, Canada

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The surface integral of ( r ⃗ − r ⃗ 0 ) . d S ⃗ (\vec r-\vec r_0).d\vec S is equal to the volume integral of ∇ . ( r ⃗ − r ⃗ 0 ) d V = 3 d V \nabla.(\vec r-\vec r_0)dV=3dV . For a sphere of radius R R , the value of this integral is 4 π R 3 4πR^3 , and the required answer is 1 4 π × 4 π R 3 = R 3 \dfrac{1}{4π}\times 4πR^3=R^3 . In this problem R = 5 R=5 and the answer is 5 3 = 125 5^3=\boxed {125} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...