I am given a question to find the volume of the solid formed by a region in the second quadrant bounded above by the curve $y=-x^3$, below by the x-axis, and on the left by the line $x=-1$, revolved around the axis $x=-2$. I intend to solve this by the cylindrical shell method.
By drawing out the curve, I have come up with a few conclusions to help me set up my integral:
- The radius of each layer of the "shell" is $x+1$, because, crudely, the interval between $x=-1$ and $x=-2$ is not bounded.
- Height of each "shell" is given by $y=-x^3$, as given by the curve itself.
- The region extends from $x=-1$ to $x=0$.
Hence, my integral should be $2\pi \int_{-1}^0 (x+1)(-x^3)dx$. However, it is now known to me that this is incorrect. May I know what is wrong in my thought process?
In general, is there any formula/procedure when calculating for volumes of solids formed by revolutions around non-standard axes (x and y-axes)?
Any advice is appreciated!
The cylindrical shell radius you are looking for is $(2 + x)$ and not ($1 + x)$.
As the rotation is of area between $x = -1$ and $x = 0$, around $x = -2$,
At $x = -1$, radius $ = 1$. At $x = 0$, radius $ = 2$.
So the correct integral should be -
$\displaystyle 2\pi \int_{-1}^{0} (2 + x) (-x^3) dx$
EDIT: added a diagram for the region (marked in yellow) that you are rotating around $x = -2$.