I am not sure how to tackle this one.
find the volume of y=x+1: x=0 and x=2, rotated about the y axis.
The previous question to this i had to work out the volume rotated about the x axis, this is where i got the points x=0 and x=2 from, but now they want me to find the volume of the solid about the yaxis. Should i sub the two x numbers into the function, to obtain y values? than work from there.
I think that's a good solution, you have also to express now $\displaystyle y$ in function of $\displaystyle x$, so: $\displaystyle x=y-1$ and then use the formula (in general):
$\displaystyle V=\pi \int_{a}^{b} [f(y)]^2dy $
as stated by Fernando, the three given lines do not enclose a region in the plane.
however, if you're speaking of the region in quadrant I bounded by the given lines as shown in the graph ...
1. rotation about the x-axis
$\displaystyle V = \pi \int_0^2 (x+1)^2 \, dx$
2. rotation about the y-axis
a. using the method of cylindrical shells w/r to x
$\displaystyle V = 2\pi \int_0^2 x(x+1) \, dx$
b. using the methods of disks and washers w/r to y (requires two integrals)
$\displaystyle V = \pi \int_0^1 2^2 \, dy + \pi \int_1^3 2^2 - (y-1)^2 \, dy$
c. finally, and probably the easiest, using geometry (cylinder volume - cone volume)
$\displaystyle V = \pi \cdot 2^2 \cdot 3 - \frac{\pi}{3} \cdot 2^2 \cdot 2$