Volume and Limit

kunny

Consider the solid K of revolution generated by rotating the curve $y=\sin x\ (0\leqq x \leqq \pi)$ around the $x$ -axis.
K is divided into the number of $2n$ parts by the number of $2n-1$ planes ,which is perpendicular to $x$ -axis,
they have all equal volume. Among the $x$ -coordinate of intersection points for these plane to intersect with $x$ -axis,

let $a_n$ be the $x$ -coordinate of intersection point which is less than $\frac{\pi}{2}$ and the closest $\frac{\pi}{2}$ .

Evaluate

$\lim_{n\to\infty} n\left(\frac{\pi}{2}-a_n\right)$

Could anyone translate this sentence in plain English. Sad

Regards!

kunny

**Posted:** Tue Feb 08, 2005 10:10 am

kunny

Can anyone solve this problem?

**Posted:** Wed Feb 09, 2005 5:49 pm

Kent Merryfield

The volume of K is

$\int_0^{\pi}\pi \sin^2x\,dx=\frac{\pi^2}2.$

We choose $a_n<\frac{\pi}2$ such that

$\int_{a_n}^{\frac{\pi}2}\pi\sin^2x\,dx=\frac{\pi^2}{4n}.$

Since the value of $\pi\sin^2x=\pi$ at $\frac{\pi}2$ , as $a_n\to\frac{\pi}2$ we have

$\int_{a_n}^{\frac{\pi}2}\pi\sin^2x\,dx=\pi\left(\frac{\pi}2-a_n\right) + o\left(\frac{\pi}2-a_n\right).$

Hence $\left(\frac{\pi}2-a_n\right)=\frac{\pi}{4n}+o\left(\frac{\pi}2-a_n\right)$ or

$n\left(\frac{\pi}2-a_n\right)=\frac{\pi}4+o(1).$

The limit is $\frac{\pi}4.$

kunny

Thank you for your reply,Kent Merryfield. Smile

**Posted:** Thu Feb 10, 2005 8:02 am