Integrating h(x,y) Over a Triangle

[Renzokuken]

1. Homework Statement
Integrate h(x,y)=yi - xj over the triangle with vertices (-2,0), (2,0), (0,2) tranversed counterclockwise
2. Homework Equations
(1-t)a+tb
3. The Attempt at a Solution
(1-t)(-2,0)+t(2,0)=(-2+4t,0)
(1-t)(2,0)+t(0,2)=(2-2t,2t)
(1-t)(0,2)+t(-2,0)=(-2t,2-2t)
integral y-x
integral (0)-(-2+4t)*4 from 0 to 1 = 0
integral (2t)*2-(2-2t)*-2 from 0 to 1 = 4
integral (2-2t)*-2-(-2t)*-2 from 0 to 1 = -4
0+4+-4=0


I don't get the right answer, the correct answer is -8

 

[mighty2000]

/3. Can you please explain where I went wrong and how to solve this problem?

it is important to approach problems systematically and accurately. In this case, it seems like you may have made a mistake in your calculations or in understanding the problem. Let's break down the problem and see where the error may have occurred.

First, let's visualize the triangle described in the problem. It has vertices at (-2,0), (2,0), and (0,2) and is being traversed counterclockwise. This means that the triangle looks like a right triangle with the right angle at the vertex (0,2) and the other two vertices at (-2,0) and (2,0), respectively.

Next, let's set up the integral. The given function, h(x,y), is yi - xj. This means that we need to integrate the y component first and then the x component. We also need to take into account the direction of traversal, which is counterclockwise. This means that we need to use the parameterization (1-t)a+tb, where t goes from 0 to 1 and a and b are the two endpoints of the line segment we are integrating over.

Using this information, we can set up the integral as follows:

∫∫h(x,y)dA = ∫∫(yi - xj)dA = ∫∫ydx - ∫∫xdy

Now, we need to set up the limits of integration. Since we are traversing the triangle counterclockwise, we can set up the following parameterization:

(1-t)(-2,0)+t(2,0) for t from 0 to 1

(1-t)(2,0)+t(0,2) for t from 0 to 1

(1-t)(0,2)+t(-2,0) for t from 0 to 1

Plugging these into the integral, we get:

∫∫ydx - ∫∫xdy = ∫0^1∫0^1 (2t)(-2+4t)dt - ∫0^1∫0^1 (2-2t)(2t)dt

Simplifying, we get:

∫∫ydx - ∫∫xdy = -4∫0