Arc length help extended to surface area and centroid.

[rock.freak667]

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}

 

[Gib Z]

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}