$y = (16-x^2)^{0.5}$ is rotated around the $x$-axis to give a sphere of radius $4$ units. Find the equation of the straight line that passes through $(-4,0)$, such that when also rotated around the $x$-axis, will create a surface that bisects the volume of the sphere.
The equation will obviously take the form $y=kx+4k$. I tried to solve for the volume bounded by the revolutions of the semicircle and the straight line being equal to exactly half the volume of the sphere. The result was a very complicated series of substitutions and expansions which yielded no valid solution (it was a $6^{\text{th}}$ degree polynomial with only $1$ positive solution of $2.27$ for $k$, which is quite obviously wrong).
My question is whether there is a simpler, perhaps more ingenious way of finding $k$.
Again, thanks in advance and all contributions welcome!
The volume of your sphere is $\,\displaystyle{\frac43\pi 4^3=\frac{256\pi}{3}}\,$ . You need a line $\,y=kx+4k\,$ , which intersects $\,y=\pm\sqrt{16-x^2}\,$ at
$$16-x^2=k^2x^2+8k^2x+16k^2\implies(k^2+1)x^2+8k^2x+16(k^2-1)=0\implies$$
$$\Delta=64k^4-64(k^4-1)=8^2\implies$$
$$x_{1,2}=\frac{-8k^2\pm8}{2(k^2+1)}=\begin{cases}\;\;\;\;\;\;-4\\{}\\-\frac{4(k^2-1)}{k^2+1}\end{cases}$$
Thus, we want
$$\pi\int\limits_{-4}^{-\frac{4(k^2-1)}{k^2+1}}(kx+4k)^2x\,dx=\left.\frac\pi{3k}(kx+4k)^3\right|_{-4}^{-\frac{4(k^2-1)}{k^2+1}}=\frac\pi{3k}\frac{8^3k^3}{(k^2+1)^3}=\frac{512k^2\pi}{3(k^2+1)^3}$$
And now we want
$$\frac{2^9k^2\pi}{3(k^2+1)^3}=\frac{2^7\pi}{3}\iff4k^2=(k^2+1)^3$$
and according to WA the resulting sixtic equation doesn't even have any real roots...