Solid of Revolution

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March 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 $y = x$ and $y = x^2$ about the y axis. The curve $y = x^2$ 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
March 20th 2009, 11:40 AM
galactus

Shells (the little cylinders you mention):

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

washers:

${\pi}\int_{0}^{1}(y-y^{2})dy$
March 20th 2009, 12:28 PM
jordanrs

thanks for the reply galactus but i dont quite understand what you mean
March 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. $2{\pi}\cdot(\text{average radius})$ $\cdot(\text{height})\cdot(\text{thickness})$

Which eventually gives us $2{\pi}\int_{a}^{b}xf(x)dx$

With washers, we have an infinite amount of circles we are finding the area

of. ${\pi}\int_{a}^{b}[f(x)]^{2}dx$ . See how it looks like the

area of a circle?. ${\pi}r^{2}$

This leads to the volume of said region.