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Jan 10, 2021 at 5:37 history edited cvgmt CC BY-SA 4.0
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Jan 10, 2021 at 4:32 comment added JDVC thanks much. The book says the answer is 2 pi. Your code looks great, but it doesn't seem to jibe with the answer key!
Jan 10, 2021 at 4:19 comment added cvgmt @JDVC This because the first and the third method cover the surface twice. The range of $\theta$ seems should be $(0,\pi)$ instead of $(0,2\pi)$
Jan 10, 2021 at 4:16 history edited cvgmt CC BY-SA 4.0
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Jan 10, 2021 at 4:03 comment added JDVC ,thanks for your help on this. But the second piece of code gives me 2 Sqrt(2) pi/3. Your first and third pieces give the 4 Sqrt(2) pi/3. Hmmm
Jan 10, 2021 at 3:59 history edited cvgmt CC BY-SA 4.0
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Jan 10, 2021 at 3:25 comment added JDVC The book says the answer here is 2pi on this second problem. So.....?
Jan 10, 2021 at 2:44 history edited cvgmt CC BY-SA 4.0
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Jan 10, 2021 at 2:18 comment added JDVC Done. With answer check, too!
Jan 10, 2021 at 2:18 vote accept JDVC
Jan 10, 2021 at 1:45 comment added cvgmt @JDVC Yes, I will consider the second question,but why my previous correct answer without up vote or only one up vote.
Jan 10, 2021 at 1:34 comment added JDVC Thanks so much! That makes sense. Really appreciated. Any thoughts on that second problem? Much obliged.
Jan 10, 2021 at 1:26 history edited cvgmt CC BY-SA 4.0
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Jan 10, 2021 at 1:18 history answered cvgmt CC BY-SA 4.0