Solving for $k$: $\pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 6\pi$

[karush]

https://www.physicsforums.com/attachments/5148
$u\left(x\right)=\sqrt{x+1}$

$\pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 2\pi$

$\pi\int_{1}^{k} u\left(x\right)^2 \,dx = 6\pi$

I don't know how to get $k$

 

[MarkFL]

The way I would set this up is:

\(\displaystyle 3\int_{-1}^{1} x+1\,dx=\int_{1}^{k} x+1\,dx\)

After you apply the FTOC, you should be able to solve for $k$. :)

 

[karush]

But aren't we dealing with volume?

 

[MarkFL]

But aren't we dealing with volume?

Yes, solids of revolution (disk method), both of which will have $\pi$ as a factor, which we can divide out. :)

 

[karush]

$u\left(x\right)=\sqrt{x+1}$

so
$\pi\int_{1}^{k} u\left(x\right) \,dx = 6\pi$

$\frac{{k}^{2}}{2}+k-\frac{3}{2}=6$

${k}^{2}+2k-15=0$

so
$k=3\ k=-5$
answer is $k=3$

 

[MarkFL]

Yes, here's my working:

\(\displaystyle 3\int_{-1}^{1} x+1\,dx=\int_{1}^{k} x+1\,dx\)

\(\displaystyle 3\left[\frac{x^2}{2}+x\right]_{-1}^1=\left[\frac{x^2}{2}+x\right]_1^k\)

\(\displaystyle 3\left(\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)\right)=\left(\left(\frac{k^2}{2}+k\right)-\left(\frac{1}{2}+1\right)\right)\)

\(\displaystyle 6=\frac{k^2}{2}+k-\frac{3}{2}\)

\(\displaystyle k^2+2k-15=0\)

\(\displaystyle (k-3)(k+5)=0\)

Since we require $1<k$, we conclude that:

\(\displaystyle k=3\).