Find the mass of the triangular region with vertices (0, 0), (2, 0), and (0, 1), with density function p( x, y ) = x^2 + y^2.
Im having trouble setting up this integral. This is what i came up with:
2 -.5x+1
S S (x^2 + y^2) dydx
0 0
Not sure if that's right tho, It would prob be better to use polar coordinates so i would be integrating:
2pi ?
S S (r^2)rdrd(theta)
0 0
Im just not sure if the bounds are right.
I think it is easier to proceed with the dx dy integration that you wrote first. I got the same answer. I did not go as far as to evalauate it, however.
To find the mass of the triangular region with the given density function, you can use a double integral. Let's go through the process step by step.
First, we need to determine the bounds of integration. Since the vertices of the triangular region are (0, 0), (2, 0), and (0, 1), we can see that the region is bound by the x-axis, the line y = 1, and the line x = 2.
Next, we need to express the density function in terms of the variables of integration. In this case, the density function is given as p(x, y) = x^2 + y^2.
Now, we can set up the double integral to find the mass. Since you mentioned using polar coordinates, we can switch to polar coordinates with the following conversions:
x = r*cos(theta)
y = r*sin(theta)
dx dy = r dr d(theta)
The double integral becomes:
∬(R) (x^2 + y^2) dy dx = ∫∫(R) (r^2)(r) dr d(theta)
Where R represents the triangular region in polar coordinates.
To determine the bounds of integration in polar coordinates, we can see that r varies from 0 to the edge of the region at any angle. At theta = 0, r varies from 0 to 2. At theta = pi/4, r varies from 0 to 2cos(theta).
Therefore, the integral becomes:
∫(0 to pi/4) ∫(0 to 2cos(theta)) (r^3) dr d(theta)
Evaluating this integral will give you the mass of the triangular region.
To find the mass of the triangular region with the given density function, you can set up the integral correctly using polar coordinates.
First, let's determine the bounds for the integral. The triangular region has vertices at (0, 0), (2, 0), and (0, 1). To find the bounds for r, we need to find the maximum value of r within this triangular region.
For the point (2, 0), the distance from the origin (0, 0) is √(2^2 + 0^2) = 2. From the point (0, 1), the distance from the origin is √(0^2 + 1^2) = 1. Hence, the maximum value of r within the triangular region is 2.
Since the region lies in the first quadrant, the angle theta should vary from 0 to π/2.
Now, let's set up the integral using polar coordinates:
∫[0 to π/2] ∫[0 to 2] (r^2) r dr dθ
Integrating with respect to r first, we get:
= ∫[0 to π/2] [(1/4)r^4] |[0 to 2] dθ
= ∫[0 to π/2] [(1/4) * 2^4 - (1/4) * 0^4] dθ
= ∫[0 to π/2] [8/4] dθ
= ∫[0 to π/2] [2] dθ
Now, integrating with respect to θ, we get:
= [2θ] |[0 to π/2]
= 2(π/2) - 2(0)
= π
Therefore, the mass of the triangular region is equal to π.