Results 1 to 5 of 5
Like Tree2Thanks
  • 2 Post By SlipEternal

Math Help - Find volume of region, double integrals?

  1. #1
    Newbie
    Joined
    Feb 2014
    From
    United States
    Posts
    6

    Find volume of region, double integrals?

    Third semester calculus; What I'm I doing wrong? I'm being mark wrong for the answer I got.

    My problem:

    Find the volume of the region under the graph of f(x,y)= 5x+y+1 and above the region y^2<=x, 0<=x<=16.

    My attempt:

    First I drew the diagram for my bounds :

    Find volume of region, double integrals?-number10_16.2.png

    From here I setup the iterated integral (double integral) with the bounds I found ( wolframalpha link ):

    int (5x+y+1)dydx, x=0 to 16, y=0 to sqrt(x) - Wolfram|Alpha

    First I integrate with respect to y by holding x constant, then I integrate with respect to x and get the answer of 6464/3 cubic units.
    Last edited by mrjust; March 30th 2014 at 10:56 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,845
    Thanks
    715

    Re: Find volume of region, double integrals?

    $y^2 \le x$ implies $-\sqrt{x} \le y \le \sqrt{x}$. Change your bounds of integration.
    Thanks from Shakarri and mrjust
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2014
    From
    United States
    Posts
    6

    Re: Find volume of region, double integrals?

    Quote Originally Posted by SlipEternal View Post
    $y^2 \le x$ implies $-\sqrt{x} \le y \le \sqrt{x}$. Change your bounds of integration.

    So you're saying that the drawing should be like so:

    Find volume of region, double integrals?-flipnumber10_16.2.png

    I don't understand the intuition behind that? So when we take the square root of y^2 <= x we obtain y<= sqrt(x) and y => -sqrt(x) ? For the latter equation the inequality sign flipped because we introduced a negative sign due to the square root being positive and negative? I think my mistake was thinking that the x-axis was bounded with y=0.
    Last edited by mrjust; March 30th 2014 at 12:06 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2014
    From
    United States
    Posts
    6

    Re: Find volume of region, double integrals?

    Thanks, I changed the bounds. And got it correct, but if someone can explain my previous question.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: Find volume of region, double integrals?

    The intuition is that in y^2 \le x, y can be either positive or negative.

    You should probably know from experience that the area y^2 \le x is the whole parabola, while y \le \sqrt{x} is only the upper half. Note that by convention, \sqrt{x} is the positive square root.

    Mechanically, when you take the positive square root of both sides of y^2 \le x, you get y \le \sqrt{x} if y is positive and -y \le \sqrt{x} if y is negative. The second inequality becomes y \ge -\sqrt{x}.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 25th 2010, 03:37 AM
  2. Double integrals to find the volume of a solid
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 23rd 2009, 03:38 AM
  3. Volume of Solid and Double Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 6th 2009, 11:36 AM
  4. Replies: 4
    Last Post: February 15th 2009, 08:10 AM
  5. Volume of ellipical region..integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 1st 2009, 05:54 PM

Search Tags


/mathhelpforum @mathhelpforum