let ( d ) be the region bounded by the positive ( x ) and ( y ) axes, and the line ( 3x + 6y = 11…

let ( d ) be the region bounded by the positive ( x ) and ( y ) axes, and the line ( 3x + 6y = 11 ).\ncompute the following integral.\n iint_{d}left(x^{2}+y^{2}\right) d a \n(use decimal notation. give your answer to two decimal places.)\n iint_{d}left(x^{2}+y^{2}\right) d a=

let ( d ) be the region bounded by the positive ( x ) and ( y ) axes, and the line ( 3x + 6y = 11 ).\ncompute the following integral.\n iint_{d}left(x^{2}+y^{2}\right) d a \n(use decimal notation. give your answer to two decimal places.)\n iint_{d}left(x^{2}+y^{2}\right) d a=

Answer

Explanation:

Step1: Determine the limits of integration

First, rewrite the line equation (3x + 6y=11) as (y=\frac{11 - 3x}{6}). The region (D) is bounded by (x = 0), (y = 0), and (y=\frac{11 - 3x}{6}). For (y = 0), we get (x=\frac{11}{3}). So, the double - integral (\iint_{D}(x^{2}+y^{2})dA=\int_{0}^{\frac{11}{3}}\int_{0}^{\frac{11 - 3x}{6}}(x^{2}+y^{2})dydx).

Step2: Integrate with respect to (y)

Integrate (\int_{0}^{\frac{11 - 3x}{6}}(x^{2}+y^{2})dy): [ \begin{align*} \int_{0}^{\frac{11 - 3x}{6}}(x^{2}+y^{2})dy&=\left[x^{2}y+\frac{y^{3}}{3}\right]_{y = 0}^{y=\frac{11 - 3x}{6}}\ &=x^{2}\cdot\frac{11 - 3x}{6}+\frac{1}{3}\left(\frac{11 - 3x}{6}\right)^{3}\ &=\frac{11x^{2}-3x^{3}}{6}+\frac{(11 - 3x)^{3}}{648} \end{align*} ]

Step3: Integrate the result with respect to (x)

Now, integrate (\int_{0}^{\frac{11}{3}}\left(\frac{11x^{2}-3x^{3}}{6}+\frac{(11 - 3x)^{3}}{648}\right)dx)

  • Integrate (\int_{0}^{\frac{11}{3}}\frac{11x^{2}-3x^{3}}{6}dx): [ \begin{align*} \int_{0}^{\frac{11}{3}}\frac{11x^{2}-3x^{3}}{6}dx&=\frac{1}{6}\left(\frac{11x^{3}}{3}-\frac{3x^{4}}{4}\right)\big|_{0}^{\frac{11}{3}}\ &=\frac{1}{6}\left(\frac{11}{3}\cdot\left(\frac{11}{3}\right)^{3}-\frac{3}{4}\cdot\left(\frac{11}{3}\right)^{4}\right)\ &=\frac{1}{6}\left(\frac{11^{4}}{81}-\frac{11^{4}}{108}\right)\ &=\frac{11^{4}}{6}\left(\frac{4 - 3}{324}\right)\ &=\frac{14641}{1944} \end{align*} ]

  • Integrate (\int_{0}^{\frac{11}{3}}\frac{(11 - 3x)^{3}}{648}dx). Let (u = 11-3x), then (du=-3dx). When (x = 0), (u = 11); when (x=\frac{11}{3}), (u = 0). [ \begin{align*} \int_{0}^{\frac{11}{3}}\frac{(11 - 3x)^{3}}{648}dx&=-\frac{1}{1944}\int_{11}^{0}u^{3}du\ &=\frac{1}{1944}\cdot\frac{u^{4}}{4}\big|_{0}^{11}\ &=\frac{14641}{7776} \end{align*} ]

  • Then (\int_{0}^{\frac{11}{3}}\left(\frac{11x^{2}-3x^{3}}{6}+\frac{(11 - 3x)^{3}}{648}\right)dx=\frac{14641}{1944}+\frac{14641}{7776}) [ \begin{align*} \frac{14641}{1944}+\frac{14641}{7776}&=\frac{14641\times4 + 14641}{7776}\ &=\frac{14641\times5}{7776}\ &\approx9.40 \end{align*} ]

Answer:

(9.40)