Find the mass of the triangular region with vertices (0, 0), (6, 0), and (0, 2), with density function ρ(x,y)=x^2+y^2
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To find the mass of the triangular region, we need to integrate the density function over the region.
The triangular region is bounded by the lines y = 0, y = 2, and x = 6.
Thus, the double integral for the mass is given by:
\[ \int_0^2 \int_0^{6-3y/2} (x^2 + y^2) \,dx \,dy \]
First, let's integrate with respect to x:
\[ \int_0^{6-3y/2} (x^2 + y^2) \,dx = \left[\frac{x^3}{3}+yx^2\right]_0^{6-3y/2} \]
\[ = \frac{(6-3y/2)^3}{3}+y(6-3y/2)^2-\frac{6^3}{3} \]
Now, we integrate with respect to y:
\[ \int_0^2 \left(\frac{(6-3y/2)^3}{3}+y(6-3y/2)^2-\frac{6^3}{3}\right) \,dy \]
Integrating, we obtain:
\[ \left[\frac{(6-3y/2)^4}{12}+\frac{(6-3y/2)^3}{3}-\frac{6^3}{3}y\right]_0^2 \]
\[ = \left(\frac{(6-3(2)/2)^4}{12}+\frac{(6-3(2)/2)^3}{3}-\frac{6^3}{3}(2)\right)-\left(\frac{(6-3(0)/2)^4}{12}+\frac{(6-3(0)/2)^3}{3}-\frac{6^3}{3}(0)\right) \]
\[ = \left(\frac{(6-3)^4}{12}+\frac{(6-3)^3}{3}-\frac{6^3}{3}(2)\right)-\left(\frac{6^4}{12}+\frac{6^3}{3}-0\right) \]
\[ = \left(\frac{3^4}{12}+\frac{3^3}{3}-2(6^3)\right)-\left(\frac{6^4}{12}+\frac{6^3}{3}\right) \]
\[ = \left(\frac{81}{12}+\frac{27}{3}-2(216)\right)-\left(\frac{1296}{12}+\frac{216}{3}\right) \]
\[ = \left(\frac{81}{12}+9-432\right)-\left(\frac{108}{12}+72\right) \]
\[ = \left(\frac{81}{12}+9-432\right)-\left(9+72\right) \]
\[ = \left(\frac{81}{12}-423\right)-81 \]
\[ = \frac{81}{12}-423-81 \]
\[ = \frac{81-423 \cdot 12 - 81 \cdot 12}{12} \]
We can see that the result is negative, which does not make sense since mass cannot be negative. There must be an error in our calculation.
Upon checking our solution, we can see that there is a mistake in the limits of integration for x. The correct limits of integration for x should be 0 to 6.
Therefore, the correct integral for the mass is:
\[ \int_0^2 \int_0^{6} (x^2 + y^2) \,dx \,dy \]
Let's redo the calculation.
Integrating with respect to x:
\[ \int_0^{6} (x^2 + y^2) \,dx = \left[\frac{x^3}{3}+yx^2\right]_0^{6} \]
\[ = \frac{(6)^3}{3}+y(6)^2-\frac{6^3}{3} \]
\[ = 72+36y-72 \]
\[ = 36y \]
Integrating with respect to y:
\[ \int_0^2 36y \,dy = \left[18y^2\right]_0^2 \]
\[ = 18(2)^2-18(0)^2 \]
\[ = 18(4)-18(0) \]
\[ = 72 \]
Therefore, the mass of the triangular region is 72.
To find the mass of the triangular region, we need to integrate the density function over the region.
First, let's find the equation of the line that connects the points (0, 0) and (6, 0). Since the y-coordinate is the same for both points, the line is simply y = 0.
Next, we need to find the limits of integration in terms of x and y. We can see that the x-coordinate varies from 0 to 6, and the y-coordinate varies from 0 to the y-coordinate of the point on the line with the same x-coordinate.
For each point on the line y = 0, the y-coordinate of the point is always 0. So, the limits of integration for y are from 0 to 0.
Therefore, our limits of integration are:
x: from 0 to 6
y: from 0 to 0
Now, we can set up the integral to find the mass:
Mass = ∬ρ(x, y) dA,
where dA is the differential area in the xy-plane. In this case, dA = dx dy.
The density function is given as ρ(x, y) = x^2 + y^2.
So, the integral becomes:
Mass = ∬(x^2 + y^2) dx dy.
Now, let's integrate with respect to x first. Since the limits of integration for y are from 0 to 0, the integral with respect to x becomes 0.
Therefore, the mass is 0.