Find the exact area of the surface obtained by rotating the curve about the x-axis?

Question

So given this curve: $$y=\sqrt{9x-18},\ \ 2 \le x \le 6$$

And using this lovely formula: $$\int2πy\sqrt{1+(\frac{dy}{dx})^2} dx$$

This is what I get for a set up:

$$\int_2^62π \sqrt{9x-18}\sqrt{(\frac{2x-3}{2x-24})} dx$$

I don't know, this set up is looks pretty weird. I think I screwed up when I was squaring (dy/dx).

Answer 1

Our function is $3\sqrt{x-2}$. We have $\frac{dy}{dx}=\frac{3}{2\sqrt{x-2}}$. Square, add $1$, and bring to a common denominator. We get $\frac{4x+1}{4(x-2)}$.

Take the square root, multiply by $(2\pi)3\sqrt{x-2}$. There is very nice cancellation.

Answer 2

You could also use $x=\frac{1}{9}(y^2+8)$ to get $\frac{dx}{dy}=\frac{2}{9}y$, so

$\displaystyle S=\int_{\sqrt{10}}^{\sqrt{46}}2\pi y\sqrt{1+\frac{4}{81}y^2} dy=\frac{2\pi}{9}\int_{\sqrt{10}}^{\sqrt{46}}y\sqrt{81+4y^2} dy$.

Now let $u=81+4y^2$ $\;$to get $\displaystyle\frac{\pi}{36}\int_{121}^{265}\sqrt{u} \;du$.