Rotating $y=9x-x^2$ along y axis

Question

Say we have a surface D in xy plane limited by x axis and $y=9x-x^2$. I need to calculate volume formed by rotating D along x axis and y axis.

For x axis it's easy. The curve is limited by (0, 9). If I did the calculations correctly the volume should be:

$$V_x = \pi\int_{b}^{a}{(f(x))^2dx} = \pi\int_{0}^{9}{(9x-x^2)^2dx} = 19683/10*\pi$$

But how would we calculate the volume for y axis? I tried to do it like this. I inverted the function which gives me two functions:

$$y = \frac{9\pm\sqrt{81-4x}}{2}$$

Now I could calculate the volume using integral formed by summing up volumes along x axis.

Then I basically moved the curves down 9 units along y axis, to make it easier to calculate the volume. The new function is now:

$$y = \frac{\sqrt{81-4x}}{2}$$

Finally I calculated the volume with:

$$V = \pi\int_{0}^{20.25}{\frac{81-4x}{4}dx} = 6561/32*\pi$$

This gives me half of the volume. And the full volume is then $V_y = 6561/16*\pi$.

This doesn't seem to be the correct answer however.

Answer 1

Using cylindrical shells, the integral is very straightforward and gives $\frac{2187\pi}{2}$. However, if you want to use the inversion method you need to consider one extra step. Firstly, solving $y=9x-x^2$ for $x$ via quadratic formula $$x=\frac{1}{2}(9\pm\sqrt{81-4y})$$ Now we have 2 curves. The formula for finding the volume of revolution in between 2 curves is $$\pi\int_a^b f(x)^2-g(x)^2dx$$ In our problem, the top curve is $$x=\frac{1}{2}(9+\sqrt{81-4y})=f(x)$$ and the lower curve is $$x=\frac{1}{2}(9-\sqrt{81-4y})=g(x)$$

The bounds of integration will be 0 and 20.25 as you calculated in your question. Plugging in $f(x)$ and $g(x)$ gives

$$\pi\int_0^{20.5} \bigg(\frac{1}{2}(9+\sqrt{81-4y}\bigg)^2 - \bigg(\frac{1}{2}(9-\sqrt{81-4y})\bigg)^2dy$$

The integral is simple enough to solve and gives the desired answer of $\frac{2187\pi}{2}$

Answer 2

Use the cylinderical shell method,

$$V=\int 2\pi r h ~dr$$

Here $r$ is the horizontal distance between the point $P$ on your curve to the y-axis, hence $r=x$, and $h$ is the height of the point $P$, which is $h=y=9x-x^2$, plug in and note the integral limits are from $0$ to $9$,

$$V=\int_0^9 2\pi x (9x-x^2) ~dx$$

You can proceed from here.