Volume Integral of xy over Triangle Area

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In summary: Yesx and 2x-2You must learn to check your own work. For ##y = 2x - 2##, when ##x = 0##, ##y = -2##. That's clearly wrong. You should check both points you have lie on that line.
  • #1
physicss
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Homework Statement
Hello, the homework statement is: calculate the 2d volume integral over the area defined by the triangle with the vertices: (0,0), (0,1)
and (2,2).
of the function xy
Relevant Equations
(0,0), (0,1)
and (2,2).
My solution is 2. would that be correct? I did use double Integrals
 
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  • #2
How about showing us your solution? Then we can tell you if it is correct and if not, where you went wrong.
 
  • #3
kuruman said:
How about showing us your solution? Then we can tell you if it is correct and if not, where you went wrong.
∬(R) xy dA = ∫(0 to 2) ∫(0 to x) xy dy dx

= ∫(0 to 2) [x^2 y/2]_0^x dx

= ∫(0 to 2) (1/2) x^3 dx

(1/8) x^4 |_0^2

= (1/8) (2^4 - 0)= 2
 
  • #4
See picture below for the triangle of interest. You need the equation for the line AB.

IntegrationTriangle.png


(Edited to fix the figure as pointed out in post #5.)
 
Last edited:
  • #5
kuruman said:
The lower limit for ##y## is zero only when ##x\leq 1##. See picture below for the triangle of interest. You need the equation for the line AB.

View attachment 328334
Point A should be ##(0, 1)##, according to the OP.
 
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  • #6
Oops. I swapped coordinates in my head.
 
  • #7
physicss said:
∬(R) xy dA = ∫(0 to 2) ∫(0 to x) xy dy dx
##x \in [0, 2], \ y \in [0, x]## represents the triangle with vertices ##(0,0), (2, 0), (2, 2)##.
 
  • #8
PeroK said:
##x \in [0, 2], \ y \in [0, x]## represents the triangle with vertices ##(0,0), (2, 0), (2, 2)##.
could I also calculate it by forming a rectangle?

∫∫R xy dA = ∫∫S u(u+v) dudv
∫∫S u(u+v) dudv = ∫0^2 ∫0^1 u(u+v) dvdu
= ∫0^2 [(u^2v/2) + (uv^2/2)]_0^1 du
= ∫0^2 (u^2/2 + u/2) du
= [(u^3/6) + (u^2/4)]_0^2
= 2.
 
  • #9
physicss said:
could I also calculate it by forming a rectangle?

∫∫R xy dA = ∫∫S u(u+v) dudv
∫∫S u(u+v) dudv = ∫0^2 ∫0^1 u(u+v) dvdu
= ∫0^2 [(u^2v/2) + (uv^2/2)]_0^1 du
= ∫0^2 (u^2/2 + u/2) du
= [(u^3/6) + (u^2/4)]_0^2
= 2.
I've no idea what you are doing there. You need to sort out the correct bounds for your integral.
 
  • #10
PS the answer is not 2.
 
  • #11
PeroK said:
PS the answer is not 2.
Hello, I recalculated it ( in a shorter way). I still get 2:

∫(0 to 2) (∫(0 to y) xy dx) dy = ∫(0 to 2) (y^3)/2 dy =2

what am I doing wrong?

thanks in advance
 
  • #12
physicss said:
Hello, I recalculated it ( in a shorter way). I still get 2:

∫(0 to 2) (∫(0 to y) xy dx) dy = ∫(0 to 2) (y^3)/2 dy =2

what am I doing wrong?

thanks in advance
Your bounds are still wrong. You are including the area above the line AB in post #4.
For a given x, what is the range of y within the triangle?
 
  • #13
physicss said:
Hello, I recalculated it ( in a shorter way). I still get 2:

∫(0 to 2) (∫(0 to y) xy dx) dy = ∫(0 to 2) (y^3)/2 dy =2

what am I doing wrong?

thanks in advance
Now you've integrated over the triangle with vertices at ##(0,0), (0, 2), (2, 2)## .

Use the figure given by @kuruman

kuruman said:
See picture below for the triangle of interest. You need the equation for the line AB.

integrationtriangle-png.png


(Edited to fix the figure as pointed out in post #5.)
What is the equation of the line passing through O and B ?

What is the equation of the line passing through A and B ?
 
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  • #14
SammyS said:
Now you've integrated over the triangle with vertices at ##(0,0), (0, 2), (2, 2)## .

Use the figure given by @kurumanWhat is the equation of the line passing through O and B ?

What is the equation of the line passing through A and B ?
x and 2x-2
 
  • #15
physicss said:
2x-2
y=2x-2? No. Try plotting that on the figure... :wink:
 
  • #16
berkeman said:
y=2x-2? No. Try plotting that on the figure... :wink:
Thanks, while writing down I swapped x and y. 0.5x+1 is AB
 
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  • #17
Great. So how does that change your integrals? :smile:
 
  • #18
berkeman said:
Great. So how does that change your integrals? :smile:
the inner integral has to go from x to 0.5x+1 xy dy and the outer from 0 to 2 dx I guess?
 
  • #19
physicss said:
the inner integral has to go from x to 0.5x+1 xy dy and the outer from 0 to 2 dx I guess?
Yes
 
  • #20
physicss said:
x and 2x-2
You must learn to check your own work. For ##y = 2x - 2##, when ##x = 0##, ##y = -2##. That's clearly wrong. You should check both points you have lie on that line.
 

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