# How do you find the area of the surface generated by rotating the curve about the y-axis #y=x^2, 0<=x<=2#?

##### 1 Answer

#### Explanation:

Since we are rotating this solid around the

The formula for the surface area of a solid generated by rotating some curve

#A=2piint_c^dg(y)sqrt(1+(g'(y))^2)dy#

We go from

We will use

#A=2piint_0^4sqrtysqrt(1+(1/(2sqrty))^2)dy#

#color(white)A=2piint_0^4sqrt(y(1+1/(4y)))dy#

#color(white)A=2piint_0^4sqrt(y+1/4)dy#

Let

#A=2piint_(1//4)^(17//4)u^(1/2)du#

#color(white)A=2pi[2/3u^(3/2)]_(1//4)^(17//4)#

#color(white)A=(4pi)/3((17/4)^(3/2)-(1/4)^(3/2))#

Note that

#A=(4pi)/3((17^(3/2)-1)/8)#

#color(white)A=pi/6(17sqrt17-1)#