## Calculus: Early Transcendentals 8th Edition

$V = \pi \times (4ln(2) - \frac{2}{3})$
Calculate the volume of the region bounded by the curves $y = ln(x)$, $y = 0$ and $x = 2$ when rotating around the y-axis. Formula to calculate volume: $\pi \times \int y^2 dx$ It's rotating around the y-axis, so inversing the formula $y = ln(x)$ yields $x = e^y$ As our boundaries are $0$ and $ln(2)$ (y crosses the x-axis at $ln(2)$), we have to subtract the volume from the line $x = 2$ rotating around the y-axis. Now, we know this we can set up the integral and calculate it: $\pi \times \int (2^2 -e^{2y}) dy = \pi \times \int (4 - e^{2y}) dy = \pi \times [4y - \frac{1}{2}e^{2y}]$ FIlling in the boundaries gives us: $\pi \times (4ln(2) - \frac{1}{2}e^{2ln(2)})-(4 \times 0 - \frac{1}{2}e^{2 \times 0} = \pi \times (4ln(2) - \frac{3}{2})$ $V = \pi \times (4ln(2) - \frac{2}{3})$