Question

The graph of the equation 2x=y^2 from A(0,0) to B(2,2) is rotated around the x axis. The surface area of the resulting solid is?

Answer 1

Write the given equation as
x = (1/2)y² or as y = √(2x)

Graph the given curve within the region (0,0) and (2,2) as shown in the figure below.
When the curve is rotated about the x-axis, an element of surface area is
dA = 2πy dx

The surface area of the resulting solid is
$A= 2\pi \int_{0}^{2} \sqrt{2x} dx = \frac{4 \sqrt{2} \pi}{3} [x^{3/2}]_{0}^{2} = \frac{16 \pi}{3}$

If the right end is considered, the extra area is π*(2²) = 4π

Answer:
The surface area of the rotated solid is (16π)/3.
If the right end is considered, it is an extra area of 4π.