Question

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

Answer 1

`pi/6(17sqrt17-1)`

Explanation:

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)`