What is the volume of the solid produced by revolving #f(x)=x^3-2x+3, x in [0,1] #around the x-axis?

1 Answer
Ken Rubenstein
Jan 25, 2016

Here's a TI screenshot of the answer:
enter image source here

As a decimal expansion, we have :

enter image source here

Explanation:

When you integrate area, you get volume. In other words, if you go backwards from area, you'll get volume.

So you just integrate with the standard disc method :

#pi* int r^2dx#.

Here, #pi*r^2# represents the area of a circle. It becomes a disc when you multiply this expression by the height dx (this product yields a 3- dimensional object) . It's like a bunch of quarters of varying radii.

Here, #r = x^3-2x+3#

That's the actual radius of your disc., varying as a function of x.

Here is a pretty pic:

enter image source here

I did this in Maple...and yes, Maple freakin' rules.

Finally, you need to "get your hands dirty" and fill in all the algebraic details as far as arriving at the antiderivative and then crunching the bounded definite integral.

That's your job.

Related questions
See all questions in Determining the Volume of a Solid of Revolution
Impact of this question
1534 views around the world
You can reuse this answer
Creative Commons License