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Currently I am studying how to integrate the area of a surface of revolution.

$$x = 1+2y^2,~~1\leq y\leq2 \textrm{ around the x axis}$$

Rewrite function in terms of x and find the derivative of $f(x)$.$$\begin{align} \frac{x-1}{2} &= y^2 \\ y &= \sqrt{\frac{x-1}{2}} \\ &= \frac{\sqrt{x-1}}{\sqrt{2}} \\ y' &= \frac{1}{\sqrt{2}} \bigg[ \sqrt{x-1} \bigg] \\ &= \frac{1}{\sqrt{2}\sqrt{x-1}} \\ &= \frac{1}{\sqrt{2x-1}} \end{align}$$

Start integrating, but I got stuck and need help. I don't know where to go from here..

$$\begin{align} SA &= 2\pi \int_{1}^{2} \sqrt{\frac{x-1}{2}} * \sqrt{1+\bigg( \frac{1}{\sqrt{2x-1}}\bigg)^2}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x-1}{2}} * \sqrt{\frac{1}{2x-1}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x-1}{2}} * \sqrt{\frac{2x-1+1}{2x-1}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{(x-1)(2x)}{2(2x-1)}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x^2-x}{2x^2-1}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x(x-1)}{2\bigg[(x+1)(x-1)\bigg]}} \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x}{2(x+1)}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x}{2x+2)}}dx \\ &= 2\pi \int_{1}^{2} \sqrt{\frac{x}{2x+2)}}dx * \bigg(\frac{\sqrt{2x+2}}{\sqrt{2x+2}} \bigg)\\ &= 2\pi \int_{1}^{2} \sqrt{2x^2+2x} ~dx \\ &= 2\pi \int_{1}^{2} \sqrt{2(x^2+x)} ~dx \\ &= 2\pi \int_{1}^{2} \sqrt{2}\sqrt{x^2+x} ~dx \\ &= 2\sqrt{2}\pi \int_{1}^{2} \sqrt{x^2+x} ~dx \\ \end{align}$$

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2 Answers

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See that you're integrating with respect to $x$ and yet your limits of integration are the $y$ limits. So you're confusing the two variables.

There are two ways to write $ds$:

$$ds = \sqrt{1+(y')^2} \; dx = \sqrt{1+(x')^2} \; dy.$$

Let's go with the second one:

$$ds = \sqrt{1+(4y)^2} \; dy = \sqrt{1+16y^2} \; dy.$$

This is the thickness of a little band of the surface cut perpendicular to the $x$-axis. The diameter of the band is $y$ (which is in terms of $y$ because we have $y$-limits and we're going to integrate with respect to $y$.) That gives:

$$\int_1^2 2\pi y \sqrt{1+16y^2} \; dy.$$

Much easier.

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The area obtained by rotating the curve parametrized by $t\to(x(t),y(t))$, $t\in[a,b]$around the $x$-axis has the area:$$ \begin{aligned} A &= \int_a^b 2\pi \text{Radius}(t)\cdot\text{ArcLength(t)}\; dt \\ &=\int_a^b 2\pi |y(t)|\;\sqrt{x'(t)^2+y'(t)^2}\; dt\ . \end{aligned} $$In our case, the parametrization is $y\to(1+2y^2,y)$ with $y\in [1,2]$, so the area is$$ \begin{aligned} A &= \int_1^2 2\pi y\sqrt{(4y)^2+1^2} \;dy \\ &= \int_1^2\pi (y^2)'\sqrt{16y^2+1} \;dy \\ &\qquad\text{substitution: }y^2=u \\ &= \int_1^4\pi \sqrt{16u+1} \;du \\ &=\left[\ -\pi\cdot\frac 23\cdot\frac 1{16}(1+16u)^{3/2}\ \right]_1^4 \\ &=\frac\pi{24}\Big(\ 65^{3/2}-17^{3/2}\ \Big)\ . \end{aligned} $$

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Computer check:

sage: var('x,y');
sage: integral( 2*pi*y*sqrt(16*y^2+1), (y,1,2) )
1/24*pi*(65*sqrt(65) - 17*sqrt(17))
sage: f(x) = sqrt((x-1)/2)
sage: integral( 2*pi*f(x)*sqrt(1 + diff(f,x)^2), (x, 3, 9) ).simplify_full()
1/24*pi*(65*sqrt(65) - 17*sqrt(17))

Note that the integral from the OP, when written in terms of $x$, has to be taken from $3=1+2\cdot \color{red}1^2$ to $9=1+2\cdot \color{red}2^2$, because $y$ varies between $\color{red}1$ and $\color{red} 2$.

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