The region R is bounded by y=4sinx^2, y=0, and sqrt(pi)/2. Find the volume when the region R is revolved about the y axis. Please help me with this one

Question

anonymous · Answer

hmm ok so think about what this means...y=0 is the x axis, and sqrt(pi)/2 is just another horizontal line, right?

anonymous · Answer

and if this region is bounded by y=4sinx^2, that means that...?

anonymous · Answer

So the integral should be taken from 0 to sqrt(pi)/2 ?

anonymous · Answer

uhm hold on let me graph this first

anonymous · Answer

oh ok so this graph looks elevated...and the sqrt is only on the pi, right? not on the 2?

anonymous · Answer

Yes

anonymous · Answer

ok so hmm ok so then the area is bound by the x axis and this one line that comes in and slices the function horizontally. and rotating arounf the y axis does, in fact, mean, that you are using the domain of 0 to the sqrt(pi)/2. ^^ good job XD now what do we do next?

anonymous · Answer

hold on is this function limited, domain-wise? because it's just infinite...

anonymous · Answer

@ganeshie8 heyyyy

ganeshie8 · Answer

lets sketch some curves :)

ganeshie8 · Answer

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