Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^2, x = 3, and y = 4 about the y-axis.

1 answer:

3 0

X=3 and y=x^2 intersect at (3,9), so your bounds are y=4 and y=9 and and . , so you use the disk method therefore you have to switch variables (even though x^2 is not invertible, because you only need the right side you can use y^(1/2)) so pi*integral( 3^2- (y^(1/2))^2 dy) making it pi*[9y-(1/2)y^2]|y=(4,9) , pi*[25/2]

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