Solids of revolution around a vertical line

Question

The region bounded by $y = \sqrt{x}$, $y = 0$ and $x = 3$ is revolved around $x = 3$. What is the volume of the shape? I am not sure how to set up the integral for this. I know how to to do it about the $x$ or $y$ axis. What changes when it is another line? Also, is there another difference if the line is not touching the function?

Answer 1

The shaded area is the region bounded by $y=\sqrt{x}$, the line $y=0$ and the line $x=3$.

When the vertical strip is revolved about the line $x=3$, a cylindrical shell is obtained. Call that point labeled as blue as the point $(x,y)$. Then it has to rotate about the line $x=3$. The curved generated is then a circle of radius $3-x$ and so its circumference is given by $2\pi(3-x)$. The area of the strip is given by $dA=ydx=\sqrt{x}dx$. Thus, the volume of the cylindrical shell element is given by $$dV=\text{circumference}\cdot \text{area of strip}=2\pi(3-x)\cdot\sqrt{x}dx.$$ Hence, the volume of the solid generated (using the Shell Method) is given by

$$V=\int_0^32\pi(3-x)\cdot\sqrt{x}dx=\frac{24\pi\sqrt{3}}{5}$$

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Alternately, using the Disk Method one can write $V$ as (see the graph below)

$$V=\int_0^{\sqrt{3}}\pi(3-y^2)^2dy=\frac{24\pi\sqrt{3}}{5}$$

Note that if we rotate the horizontal strip (see the graph below) about the line $x=3$, a disk is obtained. Call that point labeled as red as $(x,y)$. Then it has to rotate about the line $x=3$. The curved generated is a circle of radius $3-x$ and its area is $dA=\pi(3-x)^2=\pi(3-y^2)^2$. Thus, the volume of the disk element is given by$$dV=dA dx=\pi(3-y^2)dy.$$

Answer 2

The way I prefer to think about these problems is to imagine, in a sense, "how" the integral computes the volume. Imagine the solid of revolution after revolving about $x = 3$. We can imagine it being composed of many, many extremely thin disks stacked one on top of the other. Let's consider just one of the disks. It will be "thin", meaning it has a thickness of $dy$ ("$dy$" means "very tiny change in $y$" which is correct in this case because we are stacking disks bottom to top and not left to right). What would its radius be? It would be the horizontal distance between the line $x=3$ and the original curve itself. Since we are concerned with horizontal distance, it makes sense to rewrite the equation for our curve solved for $x$: \begin{align*} y = \sqrt{x} \iff x = y^2 \end{align*} So if our particular disk is at height $y$, then the original curve is $y^2$ units to the right of the $y$-axis. Therefore we can compute the radius of our tiny disk by subtracting: \begin{align*} r = 3 - y^2 \end{align*} (It turns out the order you subtract doesn't matter because in computing the volume of this tiny disk, you'll wind up squaring $r$)

Now a disk is just another name for a thin cylinder, so its volume is \begin{align*} dV &= \pi r^2 \, dy \\ &= \pi \left( 3-y^2 \right)^2 \, dy \end{align*} "$dV$" since it's only a very tiny part of the whole volume.

And if we want the volume of the entire solid, we simply sum up the tiny volumes of all of these little disks. Enter the integral sign: \begin{align*} V &= \int dV \\ &= \int_0^\sqrt{3} \pi \left( 3-y^2 \right)^2 \, dy \\ &= \pi \int_0^\sqrt{3} \left( 3-y^2 \right)^2 \, dy \end{align*} ($\sqrt{3}$ is the $y$ coordinate of where the curve intersects the line of revolution)

You also asked about what happens if the line is or is not touching the function. It turns out nothing changes. If there is a point where the curve crosses the line of revolution, it will only affect the sign of the radius $r$ when we're considering some of these tiny disks. In our case $r = 3 - y^2$ is positive since the curve lies to the left of the line $x = 3$, if we extended the curve and wanted to compute the volume of revolution further out, it would just mean $3 - y^2$ is now negative in the region above where the curve intersects $x = 3$. Happily, though, we end up squaring $r$, so it makes no difference in the end.