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superSAIyan2

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Can someone please tell me what they get as the answer for this question.

Find the volume of the solid of revolution when the arc of the curve y=sinx from x=0 to x=pi/2 is rotated about y-axis.

The answer says its (pi)^3/4 - 2pi but I keep getting 2pi

Thanks
 

Drongoski

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Show your solution resulting in 2 pi.
 
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superSAIyan2

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i did

V = integration of 2(pi)xy from x=0 to x=(pi)/2= 2(pi)xsinx

how did u get 1 - sinx?
 

superSAIyan2

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i used cylindrical shells as well. How did u get (1-sinx)?

Also i just checked the cambridge textbook and the answer says its 2pi. but in the booklet i got this question from it says your answer. Is there something special about the word 'arc of y=sinx'?
 

Drongoski

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i used cylindrical shells as well. How did u get (1-sinx)?

Also i just checked the cambridge textbook and the answer says its 2pi. but in the booklet i got this question from it says your answer. Is there something special about the word 'arc of y=sinx'?
'arc of . . .' I take to mean that section of the graph of y = sin x. "1 - sin x" is the gap between y = 1 and y = sin x; not the part between the x-axis and the graph.
 

braintic

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The problem with this question is that no REGION is specified for rotation.
I haven't done the working, but I'm guessing that one of the answers comes from rotating the region bounded by the curve and the x-axis, while the other answer comes from rotating the region bounded by the curve and the y-axis.
Specifying a curve for rotation is not sufficient.
 

superSAIyan2

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well i used the region between the x-axis and the curve. So did Drongoski use region between y-axis and curve?
 

braintic

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well i used the region between the x-axis and the curve. So did Drongoski use region between y-axis and curve?
When you add the two alternative answers, you get the volume of a cylinder of radius pi/2 and height 1, as you would expect.
So, without doing the working, I am guessing that is the case.
 

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