A thin triangle has vertices (0,0), (1,0), and (0,2). Let the density function at any given point on the triangle be represented by ρ(x,y)=1+3x+y. The mass of the triangle can be represented by p/q, where p and q are coprime positive integers. What is the value of p+q?
To find the mass of the triangle, we need to integrate the density function over the triangle.
The triangle has vertices (0,0), (1,0), and (0,2), so we can find the equation of the line connecting these points.
The equation of the line passing through (0,0) and (1,0) is y = 0. And the equation of the line passing through (0,0) and (0,2) is x = 0.
So, we have the following system of equations:
y = 0
x = 0
y = 2x
We can rewrite the equation y = 2x as x = y/2.
Next, we need to find the limits of integration for x and y.
Since x = 0 is already one of the lines bounding the triangle, the lower limit for x is 0.
For y, the lower limit is given by the equation of the line y = 2x. We substitute x = 0 into this equation to find the lower limit for y.
y = 2(0) = 0
The upper limit for y is given by the line x = y/2. We substitute x = 0 into this equation to find the upper limit for y.
x = 0 = y/2
y = 0
So, the limits of integration for x are 0 to y/2, and the limits of integration for y are 0 to 0.
Now, we can calculate the mass of the triangle by integrating the density function ρ(x,y) = 1+3x+y over the given limits.
The mass M of the triangle is given by:
M = ∬ ρ(x,y) dA
M = ∫∫ (1+3x+y) dx dy
M = ∫[0,0]∫[0,y/2] (1+3x+y) dx dy
Integrating with respect to x first, we get:
M = ∫[0,0] (x+3x^2/2 + xy) | [0,y/2] dy
M = ∫[0,0] (0+0+0) dy
Since both limits of integration for y are 0, the integral evaluates to 0.
Therefore, the mass of the triangle is 0.
The value of p+q is therefore 0+1 = 1.