Find the volume of rotation about the y-axis for the region bounded by $y=5x-x^2$, and $x^2-5x+8$

Question

Find the volume of rotation about the y-axis for the region bounded by $y=5x-x^2$, and $x^2-5x+8$

Here is an image:

Normally I can do this question, but this one is tricky because since we are rotating about the y-axis, and we are quadratic, when I solve for $x$ I get two answers, one positive and one negative.

What I mean is that if you look at the graph $5x-x^2$, then there are two halves, and if you look at the green line, that's one half of the curve (positive one), and the red half is the negative one.

Since we can see both the negative and positive sections in this graph, and both are being rotated, I wasn't sure how to create the equation that I can later integrate.

Usually in the problems I do, the negative doesn't exist, and so it's trivial. In this case I am not sure.

Answer 1

The area between two functions is $$\pi\int_{y_1}^{y_2}f^2(y)-g^2(y)dy$$ where $x=f(y)$ is right curve and $x=g(y)$ is left curve.

You have two curves which certain area from $[\dfrac{7}{4},\dfrac{25}{4}]$ and intersect in $y=4$. Without using symmetry, this interval is union of $[\dfrac{7}{4},4]$ and $[4,\dfrac{25}{4}]$. With your graph you can obtain the area \begin{align} {\bf V} &=\pi\int_\frac{7}{4}^4\left(\dfrac{5+\sqrt{4y-7}}{2}\right)^2-\left(\dfrac{5-\sqrt{4y-7}}{2}\right)^2dy \\ &+\pi\int_4^\frac{25}{4}\left(\dfrac{5+\sqrt{-4y+25}}{2}\right)^2-\left(\dfrac{5-\sqrt{-4y+25}}{2}\right)^2dy \\ &=\color{blue}{45\pi} \end{align} details left to you.

Answer 2

As your graph shows, there are no "negative" sections being rotated. The two graphs intersect at $x=1$ and $x=4$. Between these two x-values, the lower curve is $y_1=x^2-5x+8$ and the upper curve is $y_2=5x-x^2$.

Think about a small vertical strip of width $dx$ and length $y_2 - y_1$ being rotated about the y-axis. The infinitesimal volume of this rotated strip is $2\pi(y_2-y_1)dx$. You just need to integrate this from $x=1$ to $x=4$. i.e., $$V = \int_1^4 2\pi x (y_2-y_1)dx = 2\pi\int_1^4(10x^2-2x^3-8x)dx = 45 \pi$$.

Answer 3

Here I use parametric method for finding volume and also use symmetry to make the problem simpler. As the upper and lower graph about $y=4$ are the same, so we restrict ourselves to interval $[\dfrac{7}{4},4]$. For $y=5x−x^2$ we see $$y=5x−x^2=\dfrac{25}{4}-\left(x-\dfrac52\right)^2=\dfrac{25}{4}-t^2$$ so paramertic form of this curve is \begin{cases} x=t+\dfrac52,\\ y=\dfrac{25}{4}-t^2. \end{cases} where $-\dfrac32\leqslant t\leqslant0$ for left curve and $0\leqslant t\leqslant\dfrac32$ for right curve, thus $${\bf V}=\int_a^b\pi x^2\dfrac{dy}{dt}dt=2 \left(\pi \int_{-\frac{3}{2}}^0 \left(t+\frac{5}{2}\right)^22 t \, dt+\pi \int_0^{\frac{3}{2}} \left(t+\frac{5}{2}\right)^2 2 t \, dt \right)=\color{blue}{45\pi}$$

Answer 4

There seems to be some confusion here that can be clarified with a simpler approach, namely Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2\pi RA$.

In the present problem, the area is given by

$$ y_1=5x-x^2\\ y2=x^2-5x+8\\ A=\int_1^4(y_1-y_2)~dx=9\\ $$

The centroid is readily obtained because of the symmetry, to wit,

$$\mathbf{R}=(R_x,R_y)=(2.5,4)$$

And therefore, the volume of rotation about the $y$-axis is

$$V_y=2\pi R_xA=2\pi\cdot 2.5\cdot 9=45\pi$$

Similarly, the rotation about the $x$-axis is

$$V_y=2\pi R_yA=2\pi\cdot 4\cdot 9=72\pi$$

I have verified these results numerically.