It would appear it is a region enclosed by two concentric spheres.
are two spheres of respective radius 2 and 4.
Hi,
I have this problem:
the surfaces r=2 and 4, 30 degrees and 50 degrees, 20 degrees and 60 degrees identify a closed surface.
1- find the enclosed volume.
2- Find the total area of the enclosed surface. ( I think it is a typo from the teacher. It is volume not surface)
The first question is straigth forward
For the secon question I have some issues.
Do I need to take take each element of surface (in spherical coordinates)
,dS1=
dS2=
dS3=
then integrate the according the limits and the total are should be
S= S1+S2+S3
Is my reasonnig correct?
Thank you
B
Your reasoning is correct, except that the differential of the vertical arc is r*sin(phi)*d(phi).
Not r*sin(theta)*d(theta).
So your d(S1), for the concave/convex surfaces at r=2 and r=4 each, should be
[r*sin(phi)*d(phi)]*[r*d(theta)]
= (r^2)(sin(phi))(d(phi))(d(theta))
Yours is (r^2)(sin(theta))(d(theta))(d(phi)).
[Well, unless, of course, your vertical angle is theta, and your horizontal angle is phi. If that is the case, then your d(S1) is correct. It's just not the conventional way of doing it.]
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Then, d(S2), for the "horizontal" walls, should be (r*d(theta))(dr) = r*dr*d(theta)
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And, d(S3), for the "vertical" walls, should be (r*sin(phi)*d(phi))(dr) = r*sin(phi)*d(phi)*dr.