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

Since we are rotating this solid around the #y#-axis, we are concerned with the #x# distance from the #y#-axis to the function. This relation is given by #x=pmsqrty#. We're only dealing with positive #x# values, so we can reduce this to just #x=sqrty# for our case.

The formula for the surface area of a solid generated by rotating some curve #g(y)# around the #y#-axis on #yin[c,d]# is given by

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

We go from #x=0# to #x=2#, which is analogous to traveling from #y=0# to #y=4#, which is what we care about.

We will use #g(y)=sqrty#. Note that #g'(y)=1/(2sqrty)#.

#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 #u=y+1/4#. This implies that #du=dy#. We will also have to change the bounds.

#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 #4^(3/2)=8#:

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

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