# Math Help - Volume of revolution problem

1. ## Volume of revolution problem

Hi everyone, I have the following problem: What is the solid formed by rotating the region bounded by y=ln(x+1),x=0,y=2 around the y axis? The answer would be its pi/2(e^4-4e^2+7). How would one solve this problem? Would one find the intersection of y=2 and y=ln(x+1) which i got to be ((e^2)-1,ln(e^2)) and then take the integral of ln(x+1) from o to (e^2)-1 away from the area under y=2 from o to (e^2)-1 to get formula the cross section and then to find the volume integrate the formula for the cross section squared (radius) from y=0 to 2 and then times by pi?

Is this correct?

Thanks

2. ## Re: Volume of revolution problem

First of all, surely you can simplify \displaystyle \begin{align*} \ln{\left( e^2 \right)} \end{align*}...

3. ## Re: Volume of revolution problem

So my point of intersection is ((e^2)-1,2) but is the process correct?

Thanks for the reply

4. ## Re: Volume of revolution problem

First, graph the region to be revolved:

Next, compute the volume of an arbitrary disk:

$dV=\pi r^2\,dy$

$r=x=e^y-1$

Hence:

$dV=\pi (e^y-1)^2\,dy$

Finally sum up the disks by integrating:

$V=\pi\int_0^2(e^y-1)^2\,dy$

5. ## Re: Volume of revolution problem

How did you get the radius = e^y-1?

6. ## Re: Volume of revolution problem

Because the function is \displaystyle \begin{align*} y = \ln{(x + 1)} \end{align*} and the radius of your solid is given by how far horizonally you extend, so \displaystyle \begin{align*} x \end{align*}. Surely you can see that if \displaystyle \begin{align*} y = \ln{(x + 1)} \end{align*} then \displaystyle \begin{align*} x = e^y - 1 \end{align*}...

7. ## Re: Volume of revolution problem

Im sorry i cannot see that could you please explain? You extend to the x coordinate of intersection dont you which is (e^2)-1.

8. ## Re: Volume of revolution problem

No, for an arbitrary disk, the radius is found by extending from the y-axis to the x-coordinate on the logarithmic curve.

Look at the graph I provided and picture an arbitrary horizontal line in the shaded area serving as the radius of some disk.