# Arc length help extended to surface area and centroid.

Homework Helper

## Homework Statement

A curce,C, has equation $y=x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}+\lambda$ where $\lambda>0$ and $0\leq x \leq 3$

The length of C is denoted by s. Show that $s=2\sqrt{3}$
The area of the surface generated when C is rotated through one revolution about the x-axis is denoted by S. Find S in terms of $\lambda$

The y-coordinate of the centroid of the region bounded by C,the axes and the line x=3 is denoted by h. Given that $\int _{0} ^{3} y^2 dx=\frac{3}{4}+\frac{8\sqrt{3}}{5}+3\lambda^2$,show that

$$\lim_{\lambda \rightarrow \infty} \frac{S}{hs}=4\pi$$

## Homework Equations

$$ds=\sqrt{1+\left (\frac{dy}{dx} \right)^2} dx$$

Arc Length between $x=x_1 \mbox{and} x_2$ is given by

$$\int _{x_1} ^{x_2} 1 ds$$

$$S=\int _{x_1} ^{x_2} 2\pi y ds$$

$$h= \frac{\int_{x_1} ^{x_2} \frac{y^2}{2}dx}{\int _{x_1} ^{x_2} ydx}$$

## The Attempt at a Solution

I was able to show $s=2\sqrt{3}$ thanks to some help from the forum yesterday.

But I believe I am doing something wrong and I don't know where I went wrong so.

$$S=2\pi \int _{0} ^{3} (x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}+\lambda) \times \frac{1}{2}(x^{\frac{-1}{2}}+x^{\frac{1}{2}}) dx$$

$$S=2\pi \left( \int_{0} ^{3} (x^{\frac{1}{2}}-\frac{1}{3}x^{\frac{3}{2}}dx+ \lambda \int_{0} ^{3} \frac{1}{2}(x^{\frac{-1}{2}}+x^{\frac{1}{2}})dx\right)$$

$$S=2\pi \int_{0} ^{3}(1-\frac{x}{3}+x-\frac{x^2}{3})dx +4\pi \lambda \sqrt{3}$$

$$S= 2\pi \left[ \frac{-x^3}{9}+\frac{x^2}{3}+x \right]_{0} ^{3} + 4\pi \lambda \sqrt{3}$$

Which gives me $6\pi + 4\pi \lambda \sqrt{3}$

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Gib Z
Homework Helper
From your attempt, how did you get from line 1 to line 2? You didn't expand properly, perhaps you forgot about the brackets in the y term. Try expanding again, I got $$3\pi + 4\pi \lambda \sqrt{3}$$