How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y^2=4x, x=y revolved about the y-axis?

Oct 13, 2016

See below.

Explanation:

Draw a picture (sketch) of the region.
Note that the graphs intersect where $y = x$ and ${y}^{2} = 4 x$, so that is where ${x}^{2} = 4 x$.
And that happens at $\left(0 , 0\right)$ and at $\left(4 , 4\right)$.

We are asked to use shells, so we take thin representative rectangles parallel to the axis of rotation. In this case that is the $y$-axis.
The slice has solid black boundaries and the radius to the axis of rotation is shown as a dashed black line.

Since the thin part is $\mathrm{dx}$, we need both equations as functions of $x$.

${y}^{2} = 4 x$ in the first quadrant (which is where the region is) gets us $y = 2 \sqrt{x}$ for the upper function.

The lower function is $y = x$

The volume of a representative shell is $2 \pi r h \cdot \text{thickness}$

In this case, we have

radius $r = x$ (the dashed black line),

height $h = u p p e r - l o w e r = 2 \sqrt{x} - x$ and

$\text{thickness} = \mathrm{dx}$.

$x$ varies from $0$ to $4$, so the volume of the solid is:

${\int}_{0}^{4} 2 \pi x \left(2 \sqrt{x} - x\right) \mathrm{dx} = 2 \pi {\int}_{0}^{4} \left(2 {x}^{\frac{3}{2}} - {x}^{2}\right) \mathrm{dx}$

(details left to the student)

$= \frac{128 \pi}{15}$