Calculus
posted by Ian .
The region between the graphs of x=y^2 and x=4y is rotated around the line y=4.
The volume of the resulting solid is?

I assume you made a sketch.
You should have the two graphs intersecting at (0.0) and (16,4)
So we will washer discs with outer ratius of 4  x/4 and inner radius of 4  √x
area of washer disc = π(4x/4)^2 = π(4√x)^2
= π(16  2x + x^2/4  (16  8√x + x)
= π( 3x + x^2 + 8x^(1/2)
volume = π∫(3x + x^2 + 8x^(1/2) dx from 0 to 16
= π [ (3/2)x^2 + (1/3)x^3 + (16/3)x^(3/2) ] from 0 to 16
= π(384 + 4096/3 + 1024/3  0]
= π(3968/3)
check my arithmetic
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