# Area of a revolution of $x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$

I think my biggest problem here is I can not find a good way to find the square root in this problem

$$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \ \ \ \ 1 \le x \le 2$$

$$\int_1^2 2 \pi \cdot {\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}} \sqrt{1 + (1/3 (y^2 + 2)^{\frac{3}{2}})^2}$$

$$\int_1^2 2 \pi \cdot \frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \sqrt{1 + y^3 + 2y}$$

Here is where I am stuck, I have tried u-substitution and trig and I can not make this work. I have two pages of notes but I expect that they would be useless to type up.

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The problem is incompletely specified. What are you rotating around? It only comes out (sort of) nice if it is the $x$-axis. But then the formula you are using is not the right one. Also, the calculation of $\sqrt{1+\left(\frac{dx}{dy}\right)^2}$ is not correct. – André Nicolas Jun 12 '12 at 0:55

## 1 Answer

My suggestion is: take it easy and take it step by step. Don't throw everything in one expression, because it is easy to make a slip: $$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$$

$$x'=\frac{1}{2}\left(y^2+2\right)^\frac{1}{2}\cdot\underbrace{ 2y}_{\text{chain rule}}=y\left(y^2+2\right)^\frac{1}{2}$$ $$x'^2=y^2\left(y^2+2\right)$$ $$1+x'^2=1+y^2\left(y^2+2\right)=y^4+2y^2+1=\left(y^2+1\right)^2$$ Now calculating the differential it is useful to take it a step too far to simplify calculations in the end: $$ds=2\pi x\sqrt{1+x'^2}=2\pi y\left(y^2+1\right)dy=\pi\left(y^2+1\right)\cdot 2ydy=\pi\left(y^2+1\right)d\left(y^2\right)=\pi\left(y^2+1\right)d\left(y^2+1\right)$$ Where I am "pushing expressions under $d$": $f'(x)dx=d\left(f(x)+C\right)$ $$s=\pi\int_1^2\left(y^2+1\right)d\left(y^2+1\right)=\left.\frac{\pi(y^2+1)^2}{2}\right|_1^2=12\pi$$

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