Sorry for posting an obvious homework question on here, but Ive been stuck on setting this integral up for quiet some time now. The problem is:
Consider the solid obtained by rotating the region bounded by the given curves about the y-axis. y = ln x, y = 1, y=3, x = 0
I tried
$$\int_0^{20.0086}\pi ({ln(x)})^2dx$$
I used 20.0086 because ln(20.0086) = 3, but this doesnt seem correct.
On
This is made simpler if you integrate over $y$. To integrate over $y$, we use the same process we'd use for integrating over $dx$ but with switched variables.
Consider what we'd do if we wanted the volume of $f(x)$ around the x-axis with points $x=2$ and $x=5$. We'd compute $$\pi \int_2^5 (f(x))^2 dx$$ Why? Because our bounds perpendicular to the axis are at $x=2$ and $x=5$ (and 5 is the upper bound because $5 > 2$), and we're calculating using a function "parallel" to our x-axis, so we use a function of $x$.
So if we integrate over $y$, we need bounds perpendicular to our y-axis (here we will have $y=1$ and $y=3$), and the function "parallel" to the y-axis is the same function given to us in the problem, but we need to turn it into a function of $y$.
So we'd have
$y_{\text{min}} = 1$
$y_{\text{max}} = 3$
$y(x) = \log x \rightarrow x(y) = e^y$.
Then we integrate
$ \pi \int_1^3 (x(y))^2 dy = $ $$\pi \int_1^3 e^{2y} dy = \frac 12 e^2 (e^4-1) \approx 622.098$$
Hint: the inverse function of $ y=\ln x$ is $x=e^y$.
So the radius of a circle centered on the $y$ axis and limited by the curve $y=\ln x$ is $r=e^y$ and the volume of a cylinder of infinitesimal height $dy$ is $\pi r^2 dy=\pi (e^y)^2 dy$