Refer to the figure below to determine the volume generated by rotating the given region around of the specified axis:
a) $R_2$ around AB
Using the formula $A(x)=2\pi(radius)(height)$
Why my radius will be $(1-x)$ and my height will be $(1-\sqrt{x})$?
Aplying in the Volume Formula: $$=\int_{0}^{1}2\pi(1-x)(1-\sqrt{x})=\frac{7\pi}{15}$$ So my volume will be $\frac{7\pi}{15}$
I didn't understand the reason for the radius and height values.
b) $R_1$ around $OA$
c) $R_2$ around $OA$
d) $R_3$ around $OA$
e) $R_3$ around $AB$
f) $R_1$ around $BC$
Can someone explain to me how I determine the radius and height to be able to develop the letters b through e?
The radius is the distance from the solid to the axis of rotation. For part a), the axis is at $x = 1$, so the distance is $1-x$, for $x$ between $0$ and the boundary of the solid.
If you are not sure, you can do a sanity check: The closest region $R_1$ gets to the axis is a distance of $0$ when $x = 1$. Note that this agrees with the answer because $1-1 = 0$.
Let's do part (c). The axis of rotation is $y = 0$. Because our shells will now be oriented along the $y-$axis, we solve for $x$ in terms of $y$. That is, $x = y^2$, where $0 \leq y \leq 1$. The distance between $R_2$ the the axis $OA$ is $y$. The height of the shells will be $y^2$. So, this gives us $$2\pi \int_{y=0}^{1}y(y^2) dy.$$