compute \\( \\iint _ { r } 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } d a \\), where \\( r = 0,3 \\times 0,2 \\)…

compute \\( \\iint _ { r } 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } d a \\), where \\( r = 0,3 \\times 0,2 \\), in two ways. \\( \\iint _ { r } f ( x, y ) d a = \\int _ { 0 } ^ { 3 } \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } \\right) d y d x \\) \\( \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } \\right) d y \\) \\( = \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } \\right) d y + \\int _ { 0 } ^ { 2 } 4 y ^ { 2 } d y \\) first part: \\( \\left( 1 + ( x - 1 ) ^ { 2 } \\right) \\int _ { 0 } ^ { 2 } d y = \\left( 1 + ( x - 1 ) ^ { 2 } \\right) ( 2 ) \\) second part: \\( 4 \\int _ { 0 } ^ { 2 } y ^ { 2 } d y = 4 \\cdot \\frac { 8 } { 3 } = \\frac { 32 } { 3 } \\) \\( \\int _ { 0 } ^ { 3 } 2 \\left( 1 + ( x - 1 ) ^ { 2 } \\right) + \\frac { 32 } { 3 } d x \\) \\( = \\int _ { 0 } ^ { 3 } \\left( 2 + 2 ( x - 1 ) ^ { 2 } + \\frac { 32 } { 3 } \\right) d x \\) \\( = \\int _ { 0 } ^ { 3 } \\frac { 38 } { 3 } + 2 ( x - 1 ) ^ { 2 } d x \\) \\( \\int _ { 0 } ^ { 3 } \\frac { 38 } { 3 } d x = \\frac { 38 } { 3 } \\cdot 3 - \\frac { 38 } { 3 } \\cdot 0 = 38 \\) \\( \\int _ { 0 } ^ { 3 } 2 ( x - 1 ) ^ { 2 } d x = 2 \\int _ { 0 } ^ { 3 } \\left( x ^ { 2 } - 2 x + 1 \\right) d x \\) \\( = 2 \\left \\frac { x ^ { 3 } } { 3 } - \\frac { 2 x ^ { 2 } } { 2 } + x \\right _ { 0 } ^ { 3 } \\) \\( 2 ( 9 - 9 + 3 ) = 2 ( 3 ) = 6 \\) \\( 38 + 6 = 44 \\)

compute \\( \\iint _ { r } 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } d a \\), where \\( r = 0,3 \\times 0,2 \\), in two ways. \\( \\iint _ { r } f ( x, y ) d a = \\int _ { 0 } ^ { 3 } \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } \\right) d y d x \\) \\( \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } + 4 y ^ { 2 } \\right) d y \\) \\( = \\int _ { 0 } ^ { 2 } \\left( 1 + ( x - 1 ) ^ { 2 } \\right) d y + \\int _ { 0 } ^ { 2 } 4 y ^ { 2 } d y \\) first part: \\( \\left( 1 + ( x - 1 ) ^ { 2 } \\right) \\int _ { 0 } ^ { 2 } d y = \\left( 1 + ( x - 1 ) ^ { 2 } \\right) ( 2 ) \\) second part: \\( 4 \\int _ { 0 } ^ { 2 } y ^ { 2 } d y = 4 \\cdot \\frac { 8 } { 3 } = \\frac { 32 } { 3 } \\) \\( \\int _ { 0 } ^ { 3 } 2 \\left( 1 + ( x - 1 ) ^ { 2 } \\right) + \\frac { 32 } { 3 } d x \\) \\( = \\int _ { 0 } ^ { 3 } \\left( 2 + 2 ( x - 1 ) ^ { 2 } + \\frac { 32 } { 3 } \\right) d x \\) \\( = \\int _ { 0 } ^ { 3 } \\frac { 38 } { 3 } + 2 ( x - 1 ) ^ { 2 } d x \\) \\( \\int _ { 0 } ^ { 3 } \\frac { 38 } { 3 } d x = \\frac { 38 } { 3 } \\cdot 3 - \\frac { 38 } { 3 } \\cdot 0 = 38 \\) \\( \\int _ { 0 } ^ { 3 } 2 ( x - 1 ) ^ { 2 } d x = 2 \\int _ { 0 } ^ { 3 } \\left( x ^ { 2 } - 2 x + 1 \\right) d x \\) \\( = 2 \\left \\frac { x ^ { 3 } } { 3 } - \\frac { 2 x ^ { 2 } } { 2 } + x \\right _ { 0 } ^ { 3 } \\) \\( 2 ( 9 - 9 + 3 ) = 2 ( 3 ) = 6 \\) \\( 38 + 6 = 44 \\)

Answer

Explanation:

Step1: Expand the integrand

$$\iint_{R}(1+(x - 1)^{2}+4y^{2})dA=\iint_{R}1dA+\iint_{R}(x - 1)^{2}dA+\iint_{R}4y^{2}dA$$

Step2: Calculate (\iint_{R}1dA)

The region (R=[0,3]\times[0,2]) is a rectangle. Using the formula for the double - integral of (1) over a rectangle (\iint_{R}1dA=(b - a)(d - c)), where (a = 0,b = 3,c = 0,d = 2). So (\iint_{R}1dA=(3-0)\times(2 - 0)=6)

Step3: Calculate (\iint_{R}(x - 1)^{2}dA)

First, integrate with respect to (y): (\iint_{R}(x - 1)^{2}dA=\int_{0}^{3}(x - 1)^{2}dx\int_{0}^{2}dy) Since (\int_{0}^{2}dy=2), then (\int_{0}^{3}(x - 1)^{2}dx=\int_{0}^{3}(x^{2}-2x + 1)dx=\left[\frac{x^{3}}{3}-x^{2}+x\right]{0}^{3}=\frac{3^{3}}{3}-3^{2}+3=9 - 9+3 = 3) So (\iint{R}(x - 1)^{2}dA=2\times3 = 6)

Step4: Calculate (\iint_{R}4y^{2}dA)

Integrate with respect to (x) first: (\iint_{R}4y^{2}dA=4\int_{0}^{2}y^{2}dy\int_{0}^{3}dx) Since (\int_{0}^{3}dx=3) and (\int_{0}^{2}y^{2}dy=\left[\frac{y^{3}}{3}\right]{0}^{2}=\frac{8}{3}) So (\iint{R}4y^{2}dA=4\times\frac{8}{3}\times3 = 32)

Step5: Sum up the results

(\iint_{R}(1+(x - 1)^{2}+4y^{2})dA=6 + 6+32=44)

Answer:

(44)