Given: The arc of parabola `y=x^2`  from `(3,9)`  to `(4,16)`  is rotated about the y-axis. Unknown: Find the area of the resulting surface. (Show all steps in solving the integral)

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Formula for calculating the surface area of body described by arc `y=f(x)` between points `x=a` and `x=b` rotating around x-axis is

`S_x=2pi int_a^bysqrt(1+y'^2)dx`

But in your case the arc is rotating around y-axis so we must express `x` as a function of `y` . `x=sqrty`

`S_y=2piint_9^16xsqrt(1+x'^2)dy=2piint_9^16 sqrt y sqrt(1+(1/(2sqrty))^2)dy=`

`2piint_9^16sqrt y sqrt(1+1/(4y))dy=2piint_9^16sqrt(y+1/4)dy=|(t=y+1/4),(dt=dy),(t_1=9+1/4=37/4),(t_2=16+1/4=65/4)|=`

`2piint_(37/4)^(65/4)sqrt t dt=2pi(2t^(3/2))/3|_(37/4)^(65/4)=`

`(4pi)/3((65/4)^(3/2)-(37/4)^(3/2))=pi/6(65sqrt65-37sqrt37)`

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The surface area is equal to `pi/6(65sqrt65-37sqrt37)` . 

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