Find the area of the surface generated by revolving the curve...

Question

over about the y-axis

Answer 1

Revolving the curve around y-axis will generate the surface area :

S = ∫ 2 π x ds

ds = √ 1+ (dy/dx)² dx or ds = √ 1+ (dx/dy)² dy

x = y³ /2

dx/dy = 3/2 y² , (dx/dy)² = 9/4 y⁴

S = ∫ 2 π (y³ /2) √ 1+ (dx/dy)² dy

= ∫³ π y³ √ 1+ 9/4 y⁴ dy

₀

Let u =9/4 y⁴ , du = 9 y³ dy , y³ dy = 1/9 du , if y=3 ,u = 9/4 *3⁴ = 3⁶ /4, If y=0, u= 0

S = π 1/9 ∫ √ 1+ u du

3⁶ /4

= π /9 * 2/3 (1+u ) ^3/2 Ι₀ = 2 π/ 27 [ ( 1+ 3⁶ /4 ) ^3/2 - 1 ]