let ( f(x,y)=x^{2}e^{x^{2}} ) and let ( r ) be the triangle bounded by the lines ( x = 3,x=y/3 ), and ( y =…

let ( f(x,y)=x^{2}e^{x^{2}} ) and let ( r ) be the triangle bounded by the lines ( x = 3,x=y/3 ), and ( y = x ) in the ( xy - plane ).\nexpress ( iint_{r}f(x,y)da ) as a double integral in two different ways by filling in the values for the integrals below. (for one of these it will be necessary to write the double integral as a sum of two integrals, as indicated; for the other, it can be written as a single integral.\n(a) with one double integral:\n( iint_{r}f(x,y)da=int_{a}^{b}int_{c}^{d}f(x,y)dydx )\nwhere ( a = 0,b = 3,c = x,d = 3x ).\n(b) with two double integrals (you must fill in all blanks to receive any credit):\n( iint_{r}f(x,y)da=int_{a}^{b}int_{c}^{d}f(x,y)d) ( d+int_{m}^{n}int_{p}^{q}f(x,y)d) ( d )\nwhere ( a = ,b = ,c = ,d = )\nand ( m = ,n = ,p = ,q = ).\n(c) now use either approach to find the value of ( iint_{r}x^{2}e^{x^{2}}da ):\nthe value of the integral is

let ( f(x,y)=x^{2}e^{x^{2}} ) and let ( r ) be the triangle bounded by the lines ( x = 3,x=y/3 ), and ( y = x ) in the ( xy - plane ).\nexpress ( iint_{r}f(x,y)da ) as a double integral in two different ways by filling in the values for the integrals below. (for one of these it will be necessary to write the double integral as a sum of two integrals, as indicated; for the other, it can be written as a single integral.\n(a) with one double integral:\n( iint_{r}f(x,y)da=int_{a}^{b}int_{c}^{d}f(x,y)dydx )\nwhere ( a = 0,b = 3,c = x,d = 3x ).\n(b) with two double integrals (you must fill in all blanks to receive any credit):\n( iint_{r}f(x,y)da=int_{a}^{b}int_{c}^{d}f(x,y)d) ( d+int_{m}^{n}int_{p}^{q}f(x,y)d) ( d )\nwhere ( a = ,b = ,c = ,d = )\nand ( m = ,n = ,p = ,q = ).\n(c) now use either approach to find the value of ( iint_{r}x^{2}e^{x^{2}}da ):\nthe value of the integral is

Answer

Explanation:

Step1: Determine the limits for part (a)

The region (R) is bounded by (x = 3), (x=\frac{y}{3}) (or (y = 3x)), and (y=x). For the order (dydx), (x) ranges from (0) to (3). For a fixed (x), (y) ranges from (x) to (3x). So (\iint_{R}f(x,y)dA=\int_{0}^{3}\int_{x}^{3x}f(x,y)dydx), where (a = 0), (b = 3), (c=x), (d = 3x).

Step2: Determine the limits for part (b)

For the order (dxdy), we need to split the region. The intersection of (y=x) and (y = 3x) is at ((0,0)). The intersection of (y=x) and (x = 3) is ((3,3)), and the intersection of (y=3x) and (x = 3) is ((3,9)). We split the integral into two parts. For (y) from (0) to (3), (x) ranges from (\frac{y}{3}) to (y). For (y) from (3) to (9), (x) ranges from (\frac{y}{3}) to (3). So (\iint_{R}f(x,y)dA=\int_{0}^{3}\int_{\frac{y}{3}}^{y}f(x,y)dxdy+\int_{3}^{9}\int_{\frac{y}{3}}^{3}f(x,y)dxdy), where (a = 0), (b = 3), (c=\frac{y}{3}), (d = y) for the first - integral and (a = 3), (b = 9), (c=\frac{y}{3}), (d = 3) for the second - integral.

Step3: Evaluate the integral (\iint_{R}x^{2}e^{x^{2}}dA) using the order (dydx)

(\iint_{R}x^{2}e^{x^{2}}dA=\int_{0}^{3}\int_{x}^{3x}x^{2}e^{x^{2}}dydx) First, integrate with respect to (y): (\int_{0}^{3}x^{2}e^{x^{2}}\left[y\right]{y = x}^{y = 3x}dx=\int{0}^{3}x^{2}e^{x^{2}}(3x - x)dx=\int_{0}^{3}2x^{3}e^{x^{2}}dx) Let (u=x^{2}), then (du = 2xdx). When (x = 0), (u = 0); when (x = 3), (u = 9). And (x^{3}dx=\frac{1}{2}x^{2}(2xdx)) The integral becomes (\int_{0}^{9}ue^{u}du) Using integration by parts: (\int_{0}^{9}ue^{u}du=\left[ue^{u}-e^{u}\right]_{0}^{9}=(9e^{9}-e^{9})-(0 - 1)=8e^{9}+1)

Answer:

(a) (a = 0), (b = 3), (c=x), (d = 3x) (b) For (\int_{0}^{3}\int_{\frac{y}{3}}^{y}f(x,y)dxdy+\int_{3}^{9}\int_{\frac{y}{3}}^{3}f(x,y)dxdy), in the first integral (a = 0), (b = 3), (c=\frac{y}{3}), (d = y); in the second integral (a = 3), (b = 9), (c=\frac{y}{3}), (d = 3) (c) (8e^{9}+1)