Calculating Volume of Revolution: y=x-x^3 from 0 to 1 about the y-axis

[Zurtex]

Hi could you please say if this is right or wrong and if it is not where I went wrong.

I got the question:

Find the volume of the area between y=x-x^3 from 0 to 1 revolved about the y-axis.

So here are my workings:

$$V = \pi \int_0^1{x^2dy}$$
$$V = \pi \int_0^1{x^2 \frac{dy}{dx}dx}$$
$$\frac{dy}{dx} = 1 - 3x^2$$
$$V = \pi \int_0^1{x^2 \left(1 - 3x^2 \right)dx}$$
$$V = \pi \int_0^1{x^2 - 3x^4 dx}$$
$$V = \pi \left[ \frac{x^3}{3} - \frac{3x^5}{5} \right]_0^1$$
$$V = \pi \left \left( \frac{1}{3} - \frac{3}{5} \right) - \pi(0)$$
$$V = \frac{-4\pi}{15}$$

 

[Chen]

Well, the volume can't be negative, can it?

As far as I remember, the volumne you get by revolving the area below f(x) between points A and B around the X axis is:

$$V = \pi \int_a^b{f(x)^2dx}$$

Edit: argh, just noticed you're revolving it around the Y axis. Sorry.

 

[Zurtex]

Grrr, I just wrote up a load of code in Latex and it didn't post up.

I think the problem is that when I am integrating:

$$x^2 - 3x^4$$

Should I separate it into the parts above and below the x - axis, work out the area of the two parts individually?

 

[Doc Al]

I presume you want to find the volume of revolution of the area between that curve f(x) and the x-axis (between x = 0, 1), revolved around the y-axis?

Wouldn't that just be:
$$V = 2\pi \int_0^1{xf(x)dx}$$

 

[Zurtex]

Originally posted by Doc Al
I presume you want to find the volume of revolution of the area between that curve f(x) and the x-axis (between x = 0, 1), revolved around the y-axis?

Wouldn't that just be:
$$V = 2\pi \int_0^1{xf(x)dx}$$

Yes but I've never heard of that equation before, that only briefly reminds me of something I did in statistics.

 

[Chen]

Originally posted by Zurtex
Yes but I've never heard of that equation before, that only briefly reminds me of something I did in statistics.

If you take a point $$(x, y)$$ on the graph, connect it to the X axis and revolve the line around the Y axis, you would get a very thin but high donut. The perimeter of the donut is $$2\pi x$$ and its height is $$y$$, bringing its volum to $$2\pi xy$$ or $$2\pi xf(x)$$. That's how you get to:

$$V = 2\pi \int_0^1{xf(x)dx}$$

 

[Chen]

Ahh, I think I found your mistake. Since you integrate by $$dy$$, you need the take Y limits and not the X limits. You are asked to find the volume between X = 0 and X = 1, those points are (0, 1) and (1, 0). So when you integrate by Y, the definite integral needs to be from 1 to 0, and not from 0 to 1. This is why you got a negative volume, and if you solve it with the equation Zurtex posted you would in fact get 4pi/15.

 

[Zurtex]

Thanks loads :D

Knew it was just some sill mistake. Although I don't understand your volume thingy but then I am only doing A level further maths.