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?
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?
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.
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!!