Using the method of shells, what is the volume of the solid formed by rotating y = sqrt(x) from x = 0 to 4 about the y-axis?

Using the method of shells, what is the volume of the solid formed by rotating y = sqrt(x) from x = 0 to 4 about the y-axis?

• A. 128π/5
• B. 256π/5
• C. 64π/5
• D. 32π/5

Answer: (A)

Using cylindrical shells around the y-axis, take vertical slices. A slice at position x has radius equal to its distance from the y-axis, which is x, and its height is the vertical extent of the region from y = 0 up to y = √x, so height = √x. With thickness dx, the volume is V = ∫ from 0 to 4 of 2π(radius)(height) dx = ∫_0^4 2π x √x dx = 2π ∫_0^4 x^(3/2) dx. Evaluating, the antiderivative of x^(3/2) is (2/5)x^(5/2). Thus V = 2π * (2/5) [x^(5/2)]_0^4 = (4π/5) * 4^(5/2). Since 4^(5/2) = (√4)^5 = 2^5 = 32, the volume is V = (4π/5) * 32 = 128π/5.

Using cylindrical shells around the y-axis, take vertical slices. A slice at position x has radius equal to its distance from the y-axis, which is x, and its height is the vertical extent of the region from y = 0 up to y = √x, so height = √x. With thickness dx, the volume is

V = ∫ from 0 to 4 of 2π(radius)(height) dx = ∫_0^4 2π x √x dx = 2π ∫_0^4 x^(3/2) dx.

Evaluating, the antiderivative of x^(3/2) is (2/5)x^(5/2). Thus

V = 2π * (2/5) [x^(5/2)]_0^4 = (4π/5) * 4^(5/2).

Since 4^(5/2) = (√4)^5 = 2^5 = 32, the volume is

V = (4π/5) * 32 = 128π/5.