Solids of rotation: inner/outer radius of "washer" confusion

[morseterry60]

Hello everybody,

It's my first post here! I've graduated recently in biology and am teaching myself math and physics. Hope to find some help here.


Problem:
"Find the volume of the solid obtained by rotating the region bounded by the given curves and specified line"

y=x^3, y=x, x>/= 0; about the x-axis

The method calls for the inner radius to be subtracted from the outer radius and taking the integral. It seems apparent to me that the outer radius here would be y=x, because it lies above y=x^3 in the region being rotated. However both my textbook and wolfram alpha interpret it the other way around, with y=x being the outer radius. I don't understand why this would be.

My wrong answer is 3pi/18. The book says 4pi/21, and wolfram alpha says -4pi/21.

I'm getting a tad frustrated and would greatly appreciate any help.

 

[Quaid]

"Find the volume of the solid obtained by rotating the region bounded by the given curves and specified line"


y=x^3, y=x, x>/= 0; about the x-axis


The method calls for the inner radius to be subtracted from the outer radius and taking the integral. It seems apparent to me that the outer radius here would be y=x, because it lies above y=x^3 in the region being rotated. However both my textbook and wolfram alpha interpret it the other way around, with y=x being the outer radius. I don't understand why this would be.


My wrong answer is 3pi/18. The book says 4pi/21, and wolfram alpha says -4pi/21

Hi.

The equation highlighted in red is a typo, right?

The outer radius is y = x

The book's answer is correct. (Maybe you mis-entered something at Wolfram; it gave me the positive result.)

If your antiderivative is correct, then you could have made a simple arithmetic error somewhere after that; please show your antiderivative and your work on evaluating the integral.

Cheers

 

[morseterry60]

Hi.

The equation highlighted in red is a typo, right?

The outer radius is y = x

The book's answer is correct. (Maybe you mis-entered something at Wolfram; it gave me the positive result.)

If your antiderivative is correct, then you could have made a simple arithmetic error somewhere after that; please show your antiderivative and your work on evaluating the integral.

Cheers

Thanks for your response. Okay here is my work.

A(x) = pix^2 (outer radius) - pi(x^3)^2 (inner radius) = pi(x^2 - x^5)

Taking the integral:

pi * int[0 to 1] (x^2 - x^5) dx

= pi[ (x^3)/3 - (x^6)/6 ] from 0 to 1

= pi/3 - pi/6

= 3pi/18

I should have just posted my work the first time, sorry.

 

[Quaid]

A(x) = pix^2 (outer radius) - pi(x^3)^2 (inner radius) = pi(x^2 - x^5)

The expression highlighted in red is incorrect.

When raising a power to another power, we multiply the exponents.

(x^3)^2 = x^6

My guess is that you already knew this, but that your brain suffered a short-circuit. I have frequently written 2+3=6 and 2*3=5 myself, ha!

Let us know, if you need any more help.

Cheers

PS: Please don't insert English phrases into your equations. Type separate statements, if you desire to elaborate on an equation. Thank you!

 

[morseterry60]

Thanks Quaid! Yeah that was a brain fart. Thanks again for your prompt responses.