calculusintegration
I am having a little trouble figuring out how to integrate this problem.
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
$y=3(2 − x)$
$y=0$
$x=0$
This is how I think we should do it, but I am still not 100 percent sure
$$V=\pi\int_0^1 (3(2 − x))^2\,dy$$
Hint: The volume is equal to $$\int_{y=1}^2 \pi \frac{1}{y^2}\,dy.$$
Remark: You forgot the squaring part: note that a cross-section at height $y$ has area $\pi x^2$, that is, $\frac{\pi}{y^2}$.
In your solution, there is also an issue with the limits of integration. Note that the problem says that the region is bounded in particular by the lines $y=1$ and $y=2$.
The sketch may not be correct, The region of interest is bounded above by the line $y=2$, below by the line $y=1$, on the left by $x=0$ (the $y$=axis) and on the right by $y=\frac{1}{x}$.
To find out the bounds you want to find the points of intersection by which the lines you are given encloses the region. So you are given the lines $y=x, y=2, x=0$, graphically, we can see that the region is enclosed by the points $(0,0),(0,2)$ and $(2,2)$.
After finding the enclosed region, we can then revolve the region around $y=5$, and depending on which method for finding the volume you use, we integrate with respect to $x$ or $y$. (In this case, the bounds are unchanged regardless of whether you integrate with respect to $x$ or to $y$ since the solid lies in between $0$ to $2$ for either $x$ or $y$, but note that this is not always the case).
Best Answer
If $y=3(2-x)$ then $x=\frac{6-y}{3}$
Next we would the bounds for the integration. $y=0$ is given. The other is whatever $y$ value makes $x=0$. When $x=0$, $y=6$
So$$V=\pi\int_0^6 ({\frac{6-y}{3}})^2\,dy$$