$2x^{1/4}$ rotated around $y = 2x$

Question

This is the question: find the volume created by rotating $2 x^{1/4}$ around $y=2x$. I was able to define the distance between the two lines as $y/2 - (y/2 )^4$. However, I can't find the radius that is the distance between a point that is perpendicular to $2x$. How can I measure a radius against a tilted axis?

Answer 1

Distance between a point that is perpendicular to 2x

Will the Distance between a point and line formula help?

AoPS post about the distance formula. Scroll down to the distance between a point and line section.

Answer 2

The easiest way is with the cross product. Since $$\left|\left|\vec A\times\vec B\right|\right|=\left|\left|\vec A\right|\right|\,\left|\left|\vec B\right|\right|\sin\theta$$ You only need to cross up a vector from a point on the line to the point in question with a unit vector in the direction of the line get a vector whose magnitude is the perpendicular distance to the line. Along the line, $\vec r=\langle x,2x,0\rangle$, so $d\vec r=\langle1,2,0\rangle\,dx$. Then $ds=\left|\left|d\vec r\right|\right|=\sqrt5dx$, so a unit vector in the direction of the line is $$\hat r=\frac{d\vec r}{\left|\left|d\vec r\right|\right|}=\frac1{\sqrt5}\langle1,2,0\rangle$$ A point on the curve may be described as $$\vec P=\langle x,2x^{\frac14},0\rangle$$ And the origin is already a point on the line, so we need $$r=\left|\left|\hat r\times\vec P\right|\right|=\left|\left|\frac1{\sqrt5}\langle1,2,0\rangle\times\langle x,2x^{\frac14},0\rangle\right|\right|=\left|\left|\frac1{\sqrt5}\langle2x^{\frac14}-2x\rangle\right|\right|=\frac2{\sqrt5}\left(x^{\frac14}-x\right)$$ We already have our $dh=ds$, so $$V=\int\pi r^2dh=\int_0^1\pi\cdot\frac45\left(x^{\frac14}-x\right)^2\sqrt5\,dx=\frac{4\pi}{\sqrt5}\int_0^1\left(x^{\frac12}-2x^{\frac54}+x^2\right)dx=\frac{4\pi}{9\sqrt5}$$