Region bounded by $y=3x-2$, $y=\sqrt{x}$, and $x=0$ about the $y$-axis. I have been doing the washer method for all of my problems up to this one, and cannot seem to find a good resource to help guide me through the problem.
I know that I need to find $2\pi r$ and also the surface area to get the answer, but how do I go about starting?
Thank you for taking the time to help!
On
Draw a careful picture. This (for me) is essential.
Note that the problem setter has kindly chosen the numbers so that the curve $y=\sqrt{x}$ and the line $y=3x-2$ meet at the point $(1,1)$.
Now take a thin vertical strip of width "$dx$", going from $x$ to $x+dx$. Rotate the part of this strip that is below $y=\sqrt{x}$ and above $y=3x-2$ about the $y$-axis. We get a cylindrical shell of thickness $dx$, radius $x$, and "height" $\sqrt{x}-(3x-2)$. This has approximate volume $$2\pi x\left(\sqrt{x}-(3x-2)\right)\,dx.$$ "Add up" (integrate) from $x=0$ to $x=1$. Our volume is $$\int_0^1 2\pi x\left(\sqrt{x}-(3x-2)\right)\,dx.$$
Remark: Alternately, you can integrate with respect to $y$, using the washer method. You will have to split your interval of integration into two parts, $y=-2$ to $0$ and $y=0$ to $1$. The two calculations are, in this case, a little easier than with the Cylindrical Shell Method. But there are two of them.
Actually, the first integral need not be done, since it gives the volume of a cone, and you probably know a formula for that.
The region to be rotated about the $y$-axis is a near triangular region formed by $3$ vertices: $A = (0,-2)$, $B = (0,0)$, and $C = (1,1)$. The volume $V_{shell} = 2\cdot \pi rhdx$, with $r = x$, $h = \sqrt{x} - (3x-2)$, and $x$ runs from $0 \to 1$. So the volume $V = \displaystyle \int_{0}^1 2\cdot \pi x(\sqrt{x} - 3x + 2)dx$. You can take it from here.