∫[-π,π]∫[-π,π] 1+x^2 siny + y^2 sinx dy dx
= ∫[-π,π] 2π dx
...
let g(x) and h(y) be two functions:
int(c to d)int(a to b)(g(x,y)+h(x,y))dxdy=int(c to d)int(a to b)g(x,y)dxdy+int(c to d)int(a to b)h(x,y)dxdy
= ∫[-π,π] 2π dx
...
First, let's analyze the given function: (1 + x^2*sin(y) + y^2*sin(x)).
1. Symmetry of the function:
By observing the function, we can see that it contains terms involving both x and y. However, none of these terms are odd functions (functions that satisfy f(-x) = -f(x)) or even functions (functions that satisfy f(-x) = f(x)). Therefore, we cannot directly apply symmetry to simplify the integral.
2. Symmetry of the region of integration (R=[-pi, pi]x[-pi, pi]):
The given region of integration is a square in the xy-plane with sides from -π to π in both x and y directions. This region exhibits symmetry along both the x-axis and the y-axis.
Using symmetry along the x-axis:
We can split the given region of integration into two equal halves using the x-axis: R1 = [-π, π]x[0, π] and R2 = [-π, π]x[-π, 0]. Since the function and the region have symmetry along the x-axis, the integrals for R1 and R2 will be equal.
Using symmetry along the y-axis:
Similarly, we can split the given region into two equal halves using the y-axis: R3 = [0, π]x[-π, π] and R4 = [-π, 0]x[-π, π]. Again, since the function and the region have symmetry along the y-axis, the integrals for R3 and R4 will be equal.
Now, we have four identical integrals for the four regions, R1, R2, R3, and R4.
For R1 and R2:
We can rewrite the double integral as follows:
∬[R1 ∪ R2] (1 + x^2*sin(y) + y^2*sin(x)) dA = 2 * ∬R1 (1 + x^2*sin(y) + y^2*sin(x)) dA
Similarly, for R3 and R4:
∬[R3 ∪ R4] (1 + x^2*sin(y) + y^2*sin(x)) dA = 2 * ∬R3 (1 + x^2*sin(y) + y^2*sin(x)) dA
Now, we can evaluate any one of the integrals (∬R1, ∬R2, ∬R3, ∬R4), and multiply the result by 2 to obtain the final answer.
Note: Since the function does not have a simple form or an available closed-form solution, the numerical integration methods such as Monte Carlo simulation or numerical approximation techniques like Riemann sums may be used to evaluate the individual integrals (∬R1, ∬R2, ∬R3, ∬R4) after splitting the region using symmetry.
I hope this explanation helps you understand how to use symmetry to evaluate a double integral! Let me know if you have any further questions.
First, let's examine the given function: f(x, y) = 1 + x^2*sin(y) + y^2*sin(x)
Symmetry in the given region R=[-π, π] × [-π, π] implies that the integral of any odd function over this region will be zero.
Now, let's decompose the integrand into two parts: an odd function and an even function.
Odd function: g(x, y) = x^2*sin(y)
Even function: h(x, y) = 1 + y^2*sin(x)
Since the function g(x, y) is odd with respect to y, its integral over the symmetric region R will be zero:
∫∫g(x, y)dA = 0
Now, let's calculate the integral of the even function h(x, y) over the region R.
∫∫h(x, y)dA = ∫∫[1 + y^2*sin(x)]dA
We can split this double integral into two separate integrals:
= ∫∫1 dA + ∫∫y^2*sin(x)dA
The first integral ∫∫1 dA over the region R represents the area of the region, which is equal to the area of a square with side length 2π. So, it evaluates to:
= ∫∫1 dA = 1 * area(R) = 1 * (2π)^2 = 4π^2
Now, we need to evaluate the second integral:
∫∫y^2*sin(x)dA
Since the integrand y^2*sin(x) is an odd function with respect to x, its integral over the symmetric region R will also be zero:
∫∫y^2*sin(x)dA = 0
Therefore, the double integral over the region R is:
∫∫f(x, y)dA = ∫∫[g(x, y) + h(x, y)]dA
= ∫∫g(x, y)dA + ∫∫h(x, y)dA
= 0 + ∫∫h(x, y)dA
= 4Ï€^2
Hence, the value of the double integral (1+x^2*sin(y)+y^2*sin(x))dA over the region R=[-π, π] × [-π, π] is 4π^2.