I am gonna rotate this function around the y-axis, and I have to find the surface area.

Quick question: the reason we can use either definition (either derivative with respect to x or y) of arc length is because both definitions of arc length give you the... arc length unsurprisingly. We just need the arc length. It feels odd to find the surface area here with respect to dy, but find the arc length using a derivative with respect to x... but it shouldnt right?

Anyway, onto the problem:

$$y = \frac{1}{3}x^{\frac{3}{2}}$$ when $0 \leq x \leq 12$

$$\frac{dy}{dx} = \frac{1}{2} x^{\frac{1}{2}}$$

$$\frac{dy^2}{dx} = \frac{1}{4} x$$

$$SA = 2 \pi \int_0^{12} x \sqrt{1 + \frac{1}{4}x} dx$$

But now I'm stuck here...I can't usub? Can I just multiply out converting $x$ to $\sqrt{x^2}$?

  • $\begingroup$ Bring the $x$ inside the radical ($x\ge 0$ here), complete the square, and integrate. $\endgroup$ – rogerl Mar 21 at 15:57
  • $\begingroup$ Set $x=\dfrac{1}{4}\cot^2\theta$. $\endgroup$ – Paras Khosla Mar 21 at 16:21
  • $\begingroup$ Mind showing me? I'm a bit lost as to what you mean? Especially the trig sub... $\endgroup$ – Kitty Capital Mar 21 at 18:01
  • $\begingroup$ @rogerl mind showing a bit? $\endgroup$ – Kitty Capital Mar 21 at 19:16

$$\text{Let }\begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}1/4\cdot\cot^2\theta\\ -1/2\cdot \cot\theta\csc^2\theta\mathrm d\theta\end{bmatrix}$$

$$\begin{aligned}\int x\sqrt{1+\dfrac{1}{4x}}\mathrm dx&=-\dfrac{1}{8}\int\cot^2\theta \sec\theta \cot\theta\csc^2\theta\mathrm d\theta\\&=-\dfrac{1}{8}\int\dfrac{\cos^2\theta}{\sin^2\theta}\cdot\dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta}\cdot\dfrac{1}{\sin^2\theta}\mathrm d\theta\\ &=-\dfrac{1}{8}\int\cot^2\theta\csc^2\theta\mathrm d\theta\end{aligned}$$

Now simply let $u=\cot\theta\implies \mathrm du=-\csc^2\theta\mathrm d\theta$. Can you proceed?

Post OP's edit

$$\begin{aligned}\int x\sqrt{1+\dfrac{1}{4}x}\mathrm dx&=\dfrac{1}{2}\int x\sqrt{x+4}\mathrm dx\end{aligned}$$

Let $u=x+4\iff x=u-4\implies \mathrm du=\mathrm dx$

$$\begin{aligned}\dfrac{1}{2}\int x\sqrt{x+4}\mathrm dx&=\dfrac{1}{2}\int \sqrt{u}(u-4)\mathrm du\\&=\dfrac{1}{2}\int \left(u^{3/2}-4\sqrt{u}\right)\mathrm du\end{aligned}$$

Can you proceed?

  • $\begingroup$ I think the 1/4x is wrong... I made a mistake, the x should not be in the denominator? $\endgroup$ – Kitty Capital Mar 21 at 19:11
  • $\begingroup$ Either way... where did the sec ceom from? $\endgroup$ – Kitty Capital Mar 21 at 19:12
  • $\begingroup$ That is because $1+\tan^2\theta=\sec^2\theta$. $\endgroup$ – Paras Khosla Mar 21 at 19:23
  • $\begingroup$ Ohhh.. This is a pretty innovative approach. Do you see how to complete the square? I'd like to see that approach but got stuck as well. Any idea? $\endgroup$ – Kitty Capital Mar 21 at 21:20
  • 1
    $\begingroup$ You're right. With substitution. $\endgroup$ – Chris Custer Mar 22 at 13:38

Hint: Use integration by parts. We have $\int x\sqrt{1+\frac14 x}\operatorname {dx}=x(\frac83)(1+\frac14 x)^{\frac32} -(\frac83) \int (1+\frac14 x)^{\frac32}\operatorname {dx}$.

  • $\begingroup$ Where did the 8/3 come from? $\endgroup$ – Kitty Capital Mar 22 at 13:59
  • $\begingroup$ To correct for the $\frac32$ and $\frac14 $ when you differentiate. $\endgroup$ – Chris Custer Mar 22 at 14:02

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