Results 1 to 5 of 5

Math Help - Revolution problem, am I right?

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    228

    Revolution 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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    Quote Originally Posted by Kaitosan View Post
    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?

    Inner Radius: 1

    Outer Radius  e^{2x-x^2} - 1

    Use the Washer Method:

    PI INTEGRAL (OUTER^2 - INNER^2) dx with the correct bounds, which you can find by intersecting the two curves. Your two bounds are correct.

    Good luck!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    228
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    Quote Originally Posted by Kaitosan View Post
    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!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    228
    Quote Originally Posted by apcalculus View Post
    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!!
    Ah I see, thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the revolution of this problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 7th 2009, 10:33 PM
  2. Please help, revolution problem (I included diagram)
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 4th 2008, 05:34 AM
  3. Revolution problem, need help!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 1st 2008, 11:44 AM
  4. Problem Finding Volume of Revolution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 1st 2008, 06:35 AM
  5. Volume of revolution problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 16th 2008, 07:08 PM

Search Tags


/mathhelpforum @mathhelpforum