Why is this (x-1) and not (1-x)?

[flyingpig]

Homework Statement

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

y = 4x - x², y = 3 about x = 1

The answer key said

\int_{1}^{3} 2\pi (x - 1)(4x - x^2 - 3) dx

Why isn't it

\int_{1}^{3} 2\pi (1 -x)(4x - x^2 - 3) dx

(1 - x) makes so much more sense. (1 - x) essentially means (at least to me) the axis from x = x (x axis) to x = 1

 

[uart]

Because x>1 and the volume is positive.

It's a good to draw a sketch of the curves before starting a problem like this.

 

[LCKurtz]

Homework Statement

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

y = 4x - x², y = 3 about x = 1

The answer key said

\int_{1}^{3} 2\pi (x - 1)(4x - x^2 - 3) dx

Why isn't it

\int_{1}^{3} 2\pi (1 -x)(4x - x^2 - 3) dx

(1 - x) makes so much more sense. (1 - x) essentially means (at least to me) the axis from x = x (x axis) to x = 1

If x is the x value of your little dx element, the radius of revolution is the horizontal distance between x and 1. You always measure x_right - x_left to get the positive horizontal distance between them. In your problem x is on the right and 1 is on the left of your radius of revolution. So it is x - 1.

 

[flyingpig]

I don't see it, I still see my "right" is x = 1 and left as x = x

I've always interpreted the radius x as just a change of distance

 

[LCKurtz]

I don't see it, I still see my "right" is x = 1 and left as x = x

I've always interpreted the radius x as just a change of distance

In your integral, x goes from 1 to 3. x is to the right of 1.

 

[flyingpig]

Because x>1 and the volume is positive.

It's a good to draw a sketch of the curves before starting a problem like this.

I did and it looks right to me