1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving for the Volume of a Solid Using Double Integrals

  1. Mar 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the
    triangular region in the xy-plane enclosed by the lines x = 0 , y = x , and x + y = 8.


    2. Relevant equations
    V = ∫∫ Height
    Base

    3. The attempt at a solution
    I first found height, because height = z (upper) - z (lower) = (x^2 + y^2) - 0 = x^2 + y^2

    Afterwards, I began solving for the base. I know the base is enclosed by the curves y = x,
    y = x - 8, and x = 0. I proceed assuming the region is y-simple, giving me the boundaries:
    Base { x ≤ y ≤ 8-x
    and 0 ≤ x ≤ 4 }
    This sets up the double integral for me, which turns out to be:
    4 8-x
    ∫ { ∫ (x^2 + y^2) dy } dx
    0 x

    Solving this integral gives me that volume is 1024/3.
    Could someone look through my work and see if I made any errors? I'm a beginner at setting up these equations, so I suspect my integral to have a mistake. Thank you in advance.
     
  2. jcsd
  3. Mar 9, 2014 #2

    Zondrina

    User Avatar
    Homework Helper

    Hmm I'm getting ##\frac{2048}{3}##. Double your answer.

    The way I managed to get that was by starting with ##x = y## and ##x = 8 - y##.

    Then ##x = 0## implies that ##y = 0## and ##y = 8##.
     
  4. Mar 10, 2014 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your setup and answer look correct to me.
     
  5. Mar 11, 2014 #4
    Yup, thank you, I got the correct answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted