The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the y-axis, rotated around x=36.

Question

amistre64 · Answer

we should get the same results if im thinking of it correctly

amistre64 · Answer

but just in case;

think of this as a solid cylinder that we are removing the under part of the sqrt(x)

amistre64 · Answer

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amistre64 · Answer

so, cylindar itself; r=36, h=6;  pi 36^2*6 is total area of this cylindar, now we gut out the area under the sqrt(x)

anonymous · Answer

so it's like an upside down bowl

amistre64 · Answer

yes, and lets disc out the underside, i think it might integrate easier that way

anonymous · Answer

ok after I integrated I got, 18x^2 - 2/3 * x^(3/2)

amistre64 · Answer

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