# evaluation of the definite integral

Actually i am finding the arc length of curve
$24xy = y^4 + 48$ from $y = 2$ to $y = 4$
i have found the derivative with respect to $y$ and that is $$\frac{y^2}6 -\frac{x}y$$ by putting this derivative in the arc length formula i get the following

$$\int_2^4 \sqrt{1+ \left(\frac{y^2}6 -\frac{x}y\right)^2}$$

I solve this definite integral by substitution but i am failed please help me, give me some hints. I will be thankful.

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how you got this result x(y)= y^3/24 + 2/y show me – Zia ur Rahman Sep 30 '11 at 19:41

From $$24xy=y^{4}+48$$ we find $$x=\frac{1}{24}\frac{y^{4}+48}{y}.$$ Thus $$\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{8}\frac{y^{4}-16}{y^{2}}.$$

The length of the curve is $$\begin{eqnarray*} L &=&\int_{2}^{4}\sqrt{1+\left( \frac{\mathrm{d}x}{\mathrm{d}y}\right) ^{2}}\mathrm{d}y=\int_{2}^{4} \sqrt{1+\left( \frac{1}{8}\frac{y^{4}-16}{y^{2}}\right) ^{2}}\mathrm{d}y &=&\dots\end{eqnarray*}$$

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thank you very much sir – Zia ur Rahman Sep 30 '11 at 20:02
@Zia ur Rahman: You are welcome. – Américo Tavares Sep 30 '11 at 20:25

HINT It might be easier to think of $x$ as a function of $y$: $$x = \frac{y^4}{24y} + \frac{48}{24y} = \ldots (\text{simplify}),$$ where $y$ ranges from $2$ to $4$. I presume you know how to calculate the arclength when the curve is given in this form.

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how you did that, x(y)=y^3/24 + 2/y – Zia ur Rahman Sep 30 '11 at 19:42
@Zia The given equation is $24 y x = blah$. So just divide the whole thing by $24y$. Note: By $x(y)$, I mean: $x$ as a function of $y$; it does not mean $x$ multiplied by $y$. – Srivatsan Sep 30 '11 at 19:44
ok now i try to solve this whole question thank you i will contact you if i could not just after five minutes ok? – Zia ur Rahman Sep 30 '11 at 19:49
@Zia Sure. You can can post your progress either in the question itself or in the comments. – Srivatsan Sep 30 '11 at 19:51
derivative with respect to y of this function of x comes out y^2/8 - 2/y^2 now if we put this into the definite integral and suppose it u then we again need to take the derivative with respect to y that comes out du/dy = y/4 +4/y^3 now what i do? – Zia ur Rahman Sep 30 '11 at 19:59