Help with volume integration application problem using Disk or Washer Methods, revolving about x-axis, revolving about y-axis.

Question

I need to find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines: y = $\sqrt {x}$ $y=0$, and $x=3$. A) the $x-axis$ B) the $y-axis$ C) the line $x=3$ D) the line $x=9$. I've already solved part A using the Disk method, and I got $\frac92$$\pi$, which was correct, but I'm confused as to what upper bound I should be using to solve part B, and I have no idea how to start C and D.

Answer 1

When you revolve the region bound by $y = 0, x = 3, y = \sqrt x \,$, we get a paraboloid and you have already found the volume correctly.

Now for part $(b),$ when you revolve it around $Y$axis, you get horizontal circular discs stacked up starting at $y = 0$ with radius increasing as $|y|$ increases. The radius is given by

$x = y^2$ (as it is bound by curve $y = \sqrt x$)

As the outer bound of $x$ is $x = 3$, upper bound of $y = \sqrt x = \sqrt 3$. Also the lower bound as given in the question, $y = 0$.

So the desired volume $ \displaystyle V_b = \int_0^{\sqrt 3} \pi x^2 dy = \int_0^{\sqrt 3} \pi y^4 dy = \frac{9\sqrt 3 \, \pi}{5}$

Now for part $(c)$, the horizontal circular radius is

$r = 3 - x = 3 - y^2; r = 3$ when $y = 0, r = 0$ when $y = \sqrt 3$

So the desired volume $ \displaystyle V_c = \pi \int_0^{\sqrt 3} (3-y^2)^2 dy$

The same applies for part $(d)$ but the radius will be $(9 - y^2)$ and bounds will be $\sqrt 3 \le y \le 3 \, (x = 3, x = 9)$.