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
[tex]A= 2\pi \int_{0}^{2} \sqrt{2x} dx = \frac{4 \sqrt{2} \pi}{3} [x^{3/2}]_{0}^{2} = \frac{16 \pi}{3} [/tex]
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.
