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.
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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1 answer