I was learning about volumes of revolution and I have a question. Here is a curve, $y=5^\left(x-2\right)-\frac{1}{25}$, which is bounded by, $y=5$ and $x=0$, graph.
Finding the volume of the shape as it's rotated about the y-axis. First, I solved for x, because I'm rotating about the y-axis. And thus, $\textrm{radius}=\log_5\left(y+\frac{1}{25}\right)+2$, we know that the area of a given cross-section at any height of the object y is, $\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2$. The total volume of the solid is, $\int^5_0{\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2}\delta y$, which yield's, $\approx95.976$. If this is the total volume of the solid, how could I find a specific portion of the volume? If I wanted to know at exactly what height half of the volume lies I could use, $\int^x_0{\pi\left(\log_5\left(y+\frac{1}{25}\right)+2\right)^2}\delta y=\frac{\textrm{Total Volume}}{a}$, where in this case a would equal 2, (for half of the total volume), and x would equal the height at which the solid would need to be cut in-order for the volume from 0 to x would equal half of the total volume.
How could I solve for x? I'm sure there is a way to do it with the integral there, or I'm making an assumption that there must be a way to get that height after doing some operation with the original 2D curve.
I tried guessing and checking the height to, $\approx3.15193$, with that height for x the total volume is, $\approx47.98819$, so I assume there must be a way to calculate the value of x without guessing and checking or iteration (sums are fine).
An explanation including every step would be nice.
$$V=\pi \int_0^5 \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=^*\frac{2 \pi \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}\approx 95.9764$$ Therefore to know where is the $y^*$ which divides the solid in half I set (RHS is half $V$): $$\pi \int_0^t \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=\frac{\pi \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}$$ which gives the non-algebraic equation $$\frac{\pi \left(2 t+\frac{1}{25} (25 t+1) (\log (25 t+1)-2) \log (25 t+1)\right)}{\log ^2(5)}=\frac{\pi \left(125+63 \log ^2(126)-126 \log (126)\right)}{25 \log ^2(5)}$$ Setting $25t+1=z$ and simplifying $$2 z+z \log ^2(z)-2 z \log (z)=127+63 \log ^2(126)-126 \log (126)$$ Mathematica gives $$z\approx 79.7981551 \to t\approx 3.1519262$$
$^*$
$$ \int \left(\log _5\left(y+\frac{1}{25}\right)+2\right){}^2 \, dy=\int \left(\frac{\log \left(y+\frac{1}{25}\right)}{\log (5)}+2\right)^2 \, dy=$$ $$=\frac{1}{\log ^2(5)}\int \left(\log ^2\left(y+\frac{1}{25}\right)+4 \log (5) \log \left(y+\frac{1}{25}\right)+4 \log ^2(5)\right)\,dy=$$ $$F(y)=\frac{1}{25 \log ^2(5)} \left(50 y+(25 y+1) \log ^2(25 y+1)-2 (25 y+1) \log (25 y+1)\right)+C$$ Then $V=F(5)-F(0)$