How to Use the Washer Method to Find the Volume in Calculus?

[darewinder]

Hi, I need help with this calc2 question. Basically they want me to find the volume obtained by region bounded by y=2x(2-x) and x-axis. and the line of rotation is x=4.

Basically i know that i need to set up an equation based on the x variable that will give me the area and then i need to run the integral from 0 to 8 but i have no clue how to do this.

I heard that there is also shell method and a precalcus way to solve this, any idea guys?

Thanks a bunch

 

[kentm]

I am also taking Calc 2, and we just so happen to be working on the same thing. You are correct about using the Shell method for this. You use the shell method for rotation around a vertical line.

This is the formula used:

$$V=\int_a^b 2\pi(Radius)(Height)\,dx.$$

In this case your shell height is just going to be: $$-2x^2+4x$$, and your shell radius should just be $$4-x$$. Your interval will actually only be from 0 to 4 because you're dealing with the radius not the diameter. So now you should have:

$$\begin{math}V = 2\pi\int_0^4 (4-x)(-2x^2+4x)\,dx = 2\pi\int_0^4 (2x^3-8x^2-4x+16)\,dx\end{math} \begin{math}= 2\pi\left[\frac{1}{2}x^4-\frac{8}{3}x^3-2x^2+16x\right]_0^4\end{math}$$

I just learned this stuff last week or so, so you'll want to double check.

 

[darewinder]

yea you must be in my class, rutgers math 152? I think i know you.\

For the problem i know you can do it with the shell method, but they want us to do it in washer method. so i need to change the equations to set them as a variable of way so i can integrate with dy. and do you have any idea on how to do this problem using precalculus.

Thanks

 

[kentm]

The washer method is pretty simple, though it's not the best way to do this problem. The area of a washer is $$\pi(r_1^{}^2-r_1^{}^2)$$ where r1 is the radius of the outer circle, and r2 is the radius of the inner circle. Which gives us a standard function for the volume of solids where you are dealing with "washers":
$$V=\pi\int_a^b f(x)^2-g(x)^2\,dx$$ where f(x) is your upper function, and g(x) is your lower function.
Now the function $$y=2x(x-2)$$ is clearly a parabola. You say it is bound by the x axis. So we are really looking at the two points where it crosses the x-axis (0,0), and (0,2), and it's global maximum (1,2). In other words, if we integrate with respect to y, we will be integrating from 0 to 2. First you need to get the equation in terms of y which leaves two possibilities:
$$x=\frac{-4+\sqrt{16-8y}}{-4}$$
and
$$x=\frac{-4-\sqrt{16-8y}}{-4}$$
It is easier to rotate around the Y Axis, but obviously this does not produce the same results. We can modify the equations slightly so that the results are the same. $$x=\frac{-4+\sqrt{16-8y}}{-4}+2$$, and $$x=\frac{-4-\sqrt{16-8y}}{-4}+2$$. The former will be the lower function, and the latter will be the upper function. Now you can integrate:

$$V = \pi\int_0^2(\frac{-4-\sqrt{16-8y}}{-4}+2)^2-(\frac{-4+\sqrt{16-8y}}{-4}+2)^2\,dy$$

You will want to finish the integral, and check my work up to this point. It looks good to me.