calculus
posted by Taeyeon
1. Let R be the region in the first quadrant enclosed by the graphs of y=4-X , y=3x , and the y-axis.
a. Find the area of region R.
b. Find the volume of the solid formed by revolving the region R about the x-axis.
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Reiny
need the intersection ....
3x = 4-x
4x=4
x = 1
height of revolved region = radius of rotation
= 4-x - 3x = 4 - 4x
area = ∫(4-4x) dx from 0 to 1
= [4x - 2x^2] from 0 to 1
= (4 - 2) - 0
= 2
volume = π∫ (4-4x) dx from 0 to 1
= π∫(16 - 32x + 16x^2) dx from 0 to 1
= π[16x - 16x^2 + (16/3)x^3 ] from 0 to 1
= π( 16 - 16 + 16/3 - 0)
= 16π/3
check my arithmetic
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