rock.freak667

Homework Helper

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**1. Homework Statement**

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

The length of C is denoted by s. Show that [itex]s=2\sqrt{3}[/itex]

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 [itex]\lambda[/itex]

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 [itex]\int _{0} ^{3} y^2 dx=\frac{3}{4}+\frac{8\sqrt{3}}{5}+3\lambda^2[/itex],show that

[tex]\lim_{\lambda \rightarrow \infty} \frac{S}{hs}=4\pi[/tex]

**2. Homework Equations**

[tex]ds=\sqrt{1+\left (\frac{dy}{dx} \right)^2} dx[/tex]

Arc Length between [itex]x=x_1 \mbox{and} x_2[/itex] is given by

[tex]\int _{x_1} ^{x_2} 1 ds[/tex]

[tex]S=\int _{x_1} ^{x_2} 2\pi y ds[/tex]

[tex]h= \frac{\int_{x_1} ^{x_2} \frac{y^2}{2}dx}{\int _{x_1} ^{x_2} ydx}[/tex]

**3. The Attempt at a Solution**

I was able to show [itex]s=2\sqrt{3}[/itex] 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.

[tex]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[/tex]

[tex]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) [/tex]

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

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

Which gives me [itex]6\pi + 4\pi \lambda \sqrt{3}[/itex]