Question

Let R be the region in the first quadrant enclosed by the grpah of y=√(6X-4), the line y=2X and the y-axis.
a) Find the area of R.
b) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis
c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis

Answer 1

The line and the curve intersect at x=0 and x=2
Integrate sqrt(6x+4) - 2x between these two limits
∫ ((6x+4)^(1/2) - 2x) dx , from 0 to 2
let 6x+4 = u
6 dx = du
dx =(1/6) du
6x=u-4
x=-(u-4)/6
The integral becomes
(1/6) ∫ u^(1/2) du -(2/6)(1/6) ∫ (u-4) du
(1/6)∫ u^(1/2) du -(1/18) ∫ (u-4) du ----(1)
6x+4=u
when x=0, u=4
when x=2, u=16
Integrating (1)
(1/6)(2/3) u^(3/2) from 4 to 16
-(1/18) u^2/2 from 4 to 16
+ (4/18) u from 4 to 16
=6.2222-6.66667+2.66666
=2.2222