Find the volume of the solid obtained by rotating $R$ about $y=\frac{x}{5}$.

Question

Consider the region $R$ given by $$R = \{(x,y)\in\mathbb{R}^2:(x-3)^2+(y-4)^2\leq 4 \text{ and } y\leq x\}.$$ Find the volume of the solid obtained by rotating $R$ about $y=\frac{x}{5}$.

I have done some exercises involving solids of revolution but only rotated about an axis parallel to either the $x$-axis or the $y$-axis. Can you show me how I have to deal with this new kind of solids?

Answer 1

First find the intersection of the line $y = x$ with the circle. Plug in $y = x$ you get

$ (x - 3)^2 + ( x - 4)^2 = 4 $

This simplifies to,

$ 2 x^2 - 14 x + 21 = 0 $

By factoring, this becomes,

$ 2 (x - \dfrac{7}{2} )^2 + 21 - 2 \left( \dfrac{49}{4} \right) = 0 $

So,

$ (x - \dfrac{7}{2} )^2 = \dfrac{7}{4} $

The limits of $x$ are therefore $x_1, x_2$ where

$x_1 = \dfrac{7}{2} - \sqrt{\dfrac{7}{4}} $

$x_2 = \dfrac{7}{2} + \sqrt{\dfrac{7}{4}} $

and the region $R = \{ (x, y) | x_1 \le x \le x_2 , 4 - \sqrt{ 4- (x-3)^2 } \le y \le x \} \cup \{ (x,y) | x_2 \le x \le 5 , 4 - \sqrt{4 - (x-3)^2} \le y \le 4 + \sqrt{4 - (x-3)^2} \}$

If $P(x,y)$ is a point in the region, then its distance from the line $y= \dfrac{x}{5}$ is given by

$r = \dfrac{( - \dfrac{1}{5} x + y )}{\sqrt{ 1 + (1/5)^2 } } = \dfrac{ (- x + 5 y) }{\sqrt{26} } $

Now the volume integral is

$V = V_1 + V_2 $

where,

$V_1 = 2 \pi \displaystyle \int_{x = x_1}^{x_2} \int_{y = 4 - \sqrt{ 4- (x - 3)^2} }^{x} \dfrac{(-x + 5 y)}{\sqrt{26} }dy dx $

and

$V_2 = 2 \pi \displaystyle \int_{x = x_2}^5 \int_{y = 4 - \sqrt{4- (x-3)^2}}^{y = 4 + \sqrt{4 - (x - 3)^2} } \dfrac{ (-x + 5 y)}{\sqrt{26}} dy dx $

Note that the line $y = \dfrac{x}{5}$ does not intersect with the circle, because

$ (x - 3)^2 + (\dfrac{x}{5} - 4)^2 = 4 $ has a discriminant of

$ (-6 - 8/5 )^2 - 4 (9 + 16 - 4)(1 + 1/25 ) \lt 0 $

Answer 2

A rotation is in order (no pun intended). Looking at these lines, we have to rotate $x=5y$ so that it aligns with one of the axes, in this case the let's choose the $x$ axis. We can do this with trig but it would be much easier with complex numbers. In this case the direction for our line is given by the number

$$z_{axis} = 5+i$$

Since we only care about the directions the magnitude is irrelevant, so we can choose numbers which are convenient for arithmetic. Rotation in the opposite direction towards the $x$ axis will be given by its conjugate

$$\bar{z}_{axis} = 5-i$$

and now we have to rotate the circle and the other line relative to the origin

$$z_{\circ} = 3+4i$$

$$z_{-} = 1+i$$

which gives us

$$z_{\circ}' = (3+4i)(5-i)\cdot\frac{1}{\sqrt{26}} = \frac{19 + 17i}{\sqrt{26}}$$

$$z_{-}' = (1+i)(5-i) = 6 + 4i$$

$z_{\circ}'$ was divided by the magnitude of $\bar{z}_{axis}$ to keep the magnitude of the product at the same distance away from the origin as the original point. Thus our new problem is the volume of the solid generated by

$$R'= \left\{(x,y)\in\Bbb{R}^2:\left(x-\frac{19}{\sqrt{26}}\right)^2+\left(y-\frac{17}{\sqrt{26}}\right)^2 \leq 4 \text{ and } 3y \leq 2x\right\}$$

rotated around the $x$ axis. Both the original and translated circles stay in the first quadrant so there is no bookkeeping that needs to be done on self intersecting rotations.