∬(R) xy dA = ∫(0 to 2) ∫(0 to x) xy dy dxkuruman said:
Point A should be ##(0, 1)##, according to the OP.kuruman said:
##x \in [0, 2], \ y \in [0, x]## represents the triangle with vertices ##(0,0), (2, 0), (2, 2)##.physicss said:
could I also calculate it by forming a rectangle?PeroK said:
I've no idea what you are doing there. You need to sort out the correct bounds for your integral.physicss said:
Hello, I recalculated it ( in a shorter way). I still get 2:PeroK said:
Your bounds are still wrong. You are including the area above the line AB in post #4.physicss said:
Now you've integrated over the triangle with vertices at ##(0,0), (0, 2), (2, 2)## .physicss said:
What is the equation of the line passing through O and B ?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.)
x and 2x-2SammyS said:
y=2x-2? No. Try plotting that on the figure...physicss said:
Thanks, while writing down I swapped x and y. 0.5x+1 is ABberkeman said:y=2x-2? No. Try plotting that on the figure...
the inner integral has to go from x to 0.5x+1 xy dy and the outer from 0 to 2 dx I guess?berkeman said:Great. So how does that change your integrals?
Yesphysicss said:
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.physicss said:
A volume integral of xy over a triangle area refers to the double integral of the function f(x, y) = xy over a triangular region in the xy-plane. This involves calculating the integral of xy with respect to both x and y over the specified triangular domain.
To set up the integral, you need to define the limits of integration based on the vertices of the triangle. For a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3), you express the integral as a double integral: ∫∫_T xy dA. The limits for x and y will depend on the linear equations of the triangle's sides.
The steps to solve the volume integral are: 1) Determine the equations of the lines forming the triangle's sides, 2) Set up the double integral with appropriate limits for x and y, 3) Integrate with respect to y first, then x (or vice versa), 4) Evaluate the resulting integral to find the volume.
Sure, consider a triangle with vertices at (0,0), (1,0), and (0,1). The integral setup would be ∫ from 0 to 1 ∫ from 0 to 1-x (xy dy dx). First, integrate xy with respect to y from 0 to 1-x, then integrate the resulting expression with respect to x from 0 to 1.
Volume integrals of xy over triangular areas are used in various fields such as physics, engineering, and computer graphics. They are particularly useful in calculating moments of inertia, areas under surfaces, and in finite element analysis for solving partial differential equations.