Welcome to our community

Be a part of something great, join today!

What exactly is going on with my integral?

shamieh

Active member
Sep 13, 2013
539
Ex 4: Find the volume of the solid obtained by rotating the region bounded by \(\displaystyle y = x - x^2 \)and \(\displaystyle y = 0\) about the line \(\displaystyle x = 2\)


I did all the algebra, integrated etc etc. But I was thinking that the integral would be from \(\displaystyle \int^2_0\) since it says about the line \(\displaystyle x = 2\)..How do they get from \(\displaystyle \int^1_0\)? I'm almost positive it has to do with the fact that i am saying \(\displaystyle 2 - x\)...But how exactly do I know it's from \(\displaystyle \int^1_0\). I guess I see the mechanics of what is going on but I don't understand it, or maybe I don't see the mechanics at all and am lost in the dark..Any elaboration would be great. (Malthe)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The region to be revolved extends from $x=0$ to $x=1$ (sketch the graph). I recommend the shell method for this solid of revolution.
 

Petrus

Well-known member
Feb 21, 2013
739
Ex 4: Find the volume of the solid obtained by rotating the region bounded by \(\displaystyle y = x - x^2 \)and \(\displaystyle y = 0\) about the line \(\displaystyle x = 2\)


I did all the algebra, integrated etc etc. But I was thinking that the integral would be from \(\displaystyle \int^2_0\) since it says about the line \(\displaystyle x = 2\)..How do they get from \(\displaystyle \int^1_0\)? I'm almost positive it has to do with the fact that i am saying \(\displaystyle 2 - x\)...But how exactly do I know it's from \(\displaystyle \int^1_0\). I guess I see the mechanics of what is going on but I don't understand it, or maybe I don't see the mechanics at all and am lost in the dark..Any elaboration would be great. (Malthe)
Hello,
we got that \(\displaystyle y=0\) if we put that in to \(\displaystyle y=x-x^2\) we get \(\displaystyle 0=x-x^2=> x_1=0, x_2=1\)
Well like Mark Said, it is better to draw!:)
Have a nice day!
regards,
\(\displaystyle |\pi\rangle\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
We can "slice and dice" the region under consideration two ways:

Shell method: in this method we are going to integrate over $x$ using "very thin shells" (cylinders, approximately).

The center of each shell will be at $x = 2$, and the radius will vary from 2 (when $x = 0$) to 1 (when $x = 1$), because the radius of each shell is:

$r(x) = 2 - x$.

The surface area of each shell is:

$2\pi r(x)f(x)$

where $f(x) = x - x^2$, so our volume element that we integrate over will be:

$dV = 2\pi r(x)f(x)dx = 2\pi (2-x)(x - x^2)dx$

yielding the integral:

$\displaystyle V = 2\pi\int_0^1 (2x - 3x^2 + x^3)\ dx$

The solid of revolution itself looks like the "top half" of a doughnut, with a "hole" in the middle that extends from $x = 1$ to $x = 3$. We *could* integrate from 0 to 2, but after $x = 1$ we are "in the hole" and there's no additional volume to add, because we are actually integrating using the function:

$g(x) = \max(f(x),0)$

(past $x = 1$ the parabola is below the $x$-axis, and "outside the boundary").

Disk method: in this method we are actually integrating over $y$ (slicing parallel to the $x$-axis). This isn't as pretty, because now we have two radii, and "getting the radii right" is more of a chore.

So let's look at the parabola $y = x - x^2$ as if $y$ is the independent variable.

Solving for $x$ in terms of $y$ we get:

$x = \frac{1}{2} \pm \sqrt{\frac{1}{4} - y}$

Hopefully it's clear that the positive square root corresponds to the "top" branch of the parabola (when viewed with the $y$-axis horizontally), and the negative square root is the "lower" branch" (we have to consider these two branches, because otherwise we do not get a function of $y$).

The "lower branch" corresponds to the OUTER radius of our disk (well, we actually get a "washer" shape, not a disk, that is to say: an annulus). The "upper branch" corresponds to the inner radius.

The area of each washer is going to be:

$\pi(R(y)^2 - r(y)^2)$

where:

$R(y) = 2 - \left(\frac{1}{2} - \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} + \sqrt{\frac{1}{4} - y}$

and:

$r(y) = 2 - \left(\frac{1}{2} + \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} - \sqrt{\frac{1}{4} - y}$

so our volume element in this case will be:

$dV = \pi(R(y)^2 - r(y)^2)dy = 6\pi\sqrt{\frac{1}{4} - y}\ dy$

leaving us with the somewhat ugly integral to calculate:

$\displaystyle V = 6\pi\int_0^{1/4} \sqrt{\frac{1}{4} - y}\ dy$

If you get the same answer both ways, it's likely it is correct :)