Calculating Surface and Volume Integrals on a Sphere: A Problem-Based Approach

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To find the surface integral of r over a sphere of radius R centered at the origin, one can simplify the process by recognizing that the vector r aligns with the unit normal vector on the sphere's surface. The surface integral can be expressed as the integral of the vector field F = r over the surface area S of the sphere. For the volume integral of Gradient•R, it is essential to distinguish between the vector r and the variable r, as the integration involves the vector from the origin to points on the sphere. Interestingly, the first part of the problem may not require an actual integral calculation at all. This approach highlights the relationship between the vector field and the geometry of the sphere.
Noone1982
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"Find the surface integral of r over a surface of a sphere of radius and center at the origin. Also find the volume integral of Gradient•R and compare your results"


Do I just integrate r to get (1/2)r^2 and plug some limits in since the r-hats equal one?
 
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Noone1982 said:
"Find the surface integral of r over a surface of a sphere of radius and center at the origin. Also find the volume integral of Gradient•R and compare your results"


Do I just integrate r to get (1/2)r^2 and plug some limits in since the r-hats equal one?

Well, it must give you what the vector r is right?

\iint\limits_{S}\mathbf{F}dS

where S is the surface area of the sphere and F=r.
 
In xyz, r would equal R = x x^ + y y^ + z z^ but I'm just using R = r r^r to be simpler.
 
Be sure to distinguish the VECTOR r from the variable r. The vector r is the vector from the origin to the point on the sphere (of radius R?). You will actually be integrating that with the unit normal vector. Since the unit normal vector to the surface of the sphere and r are in the same direction, that is just the length of r. Hmmm, for the first part of this problem you don't actually have to do an integral at all!
 
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