# Volume Integral of xy over Triangle Area

• physicss
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.
physicss
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

How about showing us your solution? Then we can tell you if it is correct and if not, where you went wrong.

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

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

(Edited to fix the figure as pointed out in post #5.)

Last edited:
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.

kuruman
Oops. I swapped coordinates in my head.

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)##.

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.

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.

PS the answer is not 2.

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?

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?

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?

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?

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.

(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 ?

PeroK and berkeman
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

physicss said:
2x-2
y=2x-2? No. Try plotting that on the figure...

berkeman said:
y=2x-2? No. Try plotting that on the figure...
Thanks, while writing down I swapped x and y. 0.5x+1 is AB

berkeman
Great. So how does that change your integrals?

berkeman said:
Great. So how does that change your integrals?
the inner integral has to go from x to 0.5x+1 xy dy and the outer from 0 to 2 dx I guess?

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

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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