
Solid of Revolution
Im stuck on the following question as i dont understand the wording and what it is actually asking me to do.
Consider the solid generated by revolving the region in the first quadrant between the lines $\displaystyle y = x$ and $\displaystyle y = x^2 $ about the y axis. The curve $\displaystyle y = x^2$ lies on the outside of the solid.
Find the volume of this region by dividing it up into small cylinders.
any help would be most appreciated

Shells (the little cylinders you mention):
$\displaystyle 2{\pi}\int_{0}^{1}x(xx^{2})dx$
washers:
$\displaystyle {\pi}\int_{0}^{1}(yy^{2})dy$

thanks for the reply galactus but i dont quite understand what you mean

Are you completely unfamiliar with volumes of revolution?.
It's too much to explain it all from scratch.
We are revolving the region between the two graphs around the y axis.
This creates a solid with a volume. In this case, it looks kind of like a bowl.
With shells, the cross sections are parallel to the axis around which we are
revolving. therefore, we integrate w.r.t x. It is based on the area of a
cylindrical shell. $\displaystyle 2{\pi}\cdot(\text{average radius})$$\displaystyle \cdot(\text{height})\cdot(\text{thickness})$
Which eventually gives us $\displaystyle 2{\pi}\int_{a}^{b}xf(x)dx$
With washers, we have an infinite amount of circles we are finding the area
of. $\displaystyle {\pi}\int_{a}^{b}[f(x)]^{2}dx$. See how it looks like the
area of a circle?. $\displaystyle {\pi}r^{2}$
This leads to the volume of said region.