Question

please help! I have been sitting here for 4 hours trying to solve
Let R be the region bounded by y = ln x, x-axis, and x = 3.
Write, but do not solve, an integral expression that will find the volume of the solid that results from rotating R about x = -2.

Answer 1

I am not sure what to do about the rotating part because there was nothing like this problem in my lesson.

Answer 2

Rotating is incredibly simple. How to think about it is that we have a line, and then we rotate is around the x axis. What do we have? We have a circle. Now, we can all agree that this circle will have a thickness equal to the thickness of the line (let's say it's infinitely thin), which we will call dx. Now the area of the circle will be pi*r^2, where the radius is the equation in question. And then we just sum up these infinitely thin disks to get a volume.

Answer 3

I haven't done calculus in a while, so give a moment to refresh my memory on rotating around a specific x = constant line.

Answer 4

and the equation would be pi* the integral of the function. but I am not sure what boundries or anything. I have never done a problem exactly like this before.

Answer 5

The first thing you need to do is look at ln(x). ln(x) has a domain of (0, infinity). It has a finite boundary at x=0, so that will be the starting point of the integral. We are then told we are bounded by x =3, which gives us our second boundary of x =3. So our integral will be from 0 to 3.

NOTE: We only have to set up this integral, so do not worry about solving it.