I'm trying to calculate using the disk/washer method and the shell method of the volume of revolution bounded by the lines y = 0, y = x, and the circle x^2+y^2 = 1 . Rotated about the x-axis.
For the Disk/Washer method, I set it up as V= pi * integral from 0 to 1 * x^2 dx = pi/3
Confused on how to set it up with the Cylindrical Shell method to get the same Volume.
In the disk method, the radius function $r(x)$ is $x$ is $0 \le x \le \frac{1}{\sqrt{2}}$ and is $\sqrt{1 - x^2}$ if $\frac{1}{\sqrt{2}} \le x \le 1$. So via the disk method, $$V = \pi \int_0^1 r(x)^2\, dx = \pi \int_0^{1/\sqrt{2}} x^2\, dx + \pi \int_{1/\sqrt{2}}^1 (1 - x^2)\, dx$$ To apply the shell method, take cross-sections parallel to the axis of revolution (in this case, the $x$-axis). They would be horizontal strips; if a strip is a distance $y$ above the $x$-axis, its length is $\sqrt{1 - y^2} - y$. By the shell method the volume integral is then
$$V = 2\pi \int_0^1 y\left(\sqrt{1-y^2} - y\right)\, dy$$