Question

Let R be the region in the first quadrant enclosed by the graph y=(square root(6x+4)), the line y=2x, and the y-axis. . . .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 two boundary curves y = √(6x + 4) and y = 2x meet at

√(6x + 4) = 2x

6x + 4 = 4x²

2x² - 3x - 2 = 0

(x - 2) (2x + 1) = 0

⇒   x = -1/2 and x = 2

R is bounded to the left by the y-axis (x = 0), so R is the set

R = {(x, y) : 0 ≤ x ≤ 2 and 2x ≤ y ≤ √(6x + 4)}

Using the shell method, the volume is made up of cylindrical shells of radius x and height √(6x + 4) - 2x. So each shell of thickness ∆x contributes a volume of

2π (radius) (height) ∆x = 2π x (√(6x + 4) - 2x) ∆x

and as we let ∆x approach zero, the total volume of the solid is given by the definite integral

$\displaystyle \boxed{2\pi \int_0^2 x \left(\sqrt{6x + 4} - 2x\right) \, dx}$