Volume Integral of xy over Triangle Area

[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

 

[kuruman]

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

 

[physicss]

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

 

[kuruman]

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.)

 

[PeroK]

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.

 

[PeroK]

∬(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 $(0,0), (2, 0), (2, 2)$.

 

[physicss]

$x \in [0, 2], \ y \in [0, x]$ 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.

 

[PeroK]

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.

 

[PeroK]

PS the answer is not 2.

 

[physicss]

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

 

[haruspex]

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?

 

[SammyS]

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

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 ?

 

[physicss]

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

 

[berkeman]

2x-2

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

 

[physicss]

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?

 

[physicss]

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?

 

[haruspex]

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

 

[PeroK]

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.