1. ## Volume...finding k

Let $k$ be a real number between 0 and $\pi$. Let $R$ be the region in the first quadrant bounded by the x-axis, the line $x=k$ and the curve $y=sin\:x\0\le" alt="y=sin\:x\0\le" /> $\:x\:$ $\le$ $k$). If the volume of the solid generated by revolving the region $R$ about the x-axis is $\frac{\pi}{24}(5\pi-3)$, find the value of $k$.

Thanks!

2. Use the washers method and find k.

${\pi}\int_{0}^{k}sin^{2}(x)dx=\frac{(k-sin(k)cos(k)){\pi}}{2}$

$\frac{(k-sin(k)cos(k)){\pi}}{2}=\frac{\pi}{24}(5\pi-3)$

$k-sin(k)cos(k)=\frac{1}{12}(5\pi-3)$

$2k-sin(2k)=\frac{1}{6}(5\pi-3)$

Can you solve for k?. Try Newton'e method if all else fails.

3. Originally Posted by galactus
Use the washers method and find k.

${\pi}\int_{0}^{k}sin^{2}(x)dx=\frac{(k-sin(k)cos(k)){\pi}}{2}$

$\frac{(k-sin(k)cos(k)){\pi}}{2}=\frac{\pi}{24}(5\pi-3)$

$k-sin(k)cos(k)=\frac{1}{12}(5\pi-3)$

$2k-sin(2k)=\frac{1}{6}(5\pi-3)$

Can you solve for k?. Try Newton'e method if all else fails.
how would you isolated for k, or dont you, is it just inspection?

4. This one converges rather quickly, so Newton's method is a decent choice.

Let your initial guess be 1.5.

$f'(k)=2-2cos(2k)$

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$

Let k=1.5:

$k-\frac{2k-sin(2k)-\frac{1}{6}(5\pi-3)}{2-2cos(2k)}=1.31384700816$

Now, do it again with the answer we got above and we get 1.30900315524

Again: 1.30899693901

That's good enough.