Let R be the region bounded by the graph of y = e^(2x-x^2) and the horizontal line y = 2. What is the volume of R revolved about y = 1?

I gave my answer as pi{[e^(2x-x^2)]^2 - [2-e^(2x-x^2)]^2 dx with limits from 0.446 to 1.554.

Am I right?

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- April 28th 2009, 06:32 AMKaitosanRevolution problem, am I right?
Let R be the region bounded by the graph of y = e^(2x-x^2) and the horizontal line y = 2. What is the volume of R revolved about y = 1?

I gave my answer as pi{[e^(2x-x^2)]^2 - [2-e^(2x-x^2)]^2 dx with limits from 0.446 to 1.554.

Am I right? - April 28th 2009, 06:53 AMapcalculus
- April 28th 2009, 07:30 AMKaitosan
I don't get it. It looks like you're including the area from y=e^(2x-x^2) to y=1, while in reality R is bounded by y=2 and y=e^(2x-x^2). Or maybe the problem allows for you to do both ways, since it looks like both ways satisfies the condition of being revolved around y=1.

- April 28th 2009, 10:15 AMapcalculus

No I am not. including it. The distance from the rotational axis to the nearest boundary on the region is 1 unit (the distance from y=1 to y=2) at every level x. Call this the inner radius r(x).

The distance from the rotational axis to the outer boundary of the region is the y-coordinate of your exponential function minus 1, due the the rotational axis being y=1, one unit above the horizontal axis. Call this R(x).

The washer method is a difference between two disks:

INTEGRAL (Pi R^2) - INTEGRAL (Pi r^2)

= Pi INTEGRAL (R^2 - r^2).

I have posted handout of the washer and disk methods on my website.

A direct link of this document: ap calculus disk washer methods - Fullscreen

Good luck!! - April 28th 2009, 10:39 AMKaitosan