# Solid of Revolution

• Mar 20th 2009, 09:51 AM
jordanrs
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
• Mar 20th 2009, 11:40 AM
galactus
Shells (the little cylinders you mention):

$\displaystyle 2{\pi}\int_{0}^{1}x(x-x^{2})dx$

washers:

$\displaystyle {\pi}\int_{0}^{1}(y-y^{2})dy$
• Mar 20th 2009, 12:28 PM
jordanrs
thanks for the reply galactus but i dont quite understand what you mean
• Mar 20th 2009, 01:01 PM
galactus
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.