use symmetry to evaluate the double integral (1+x^2siny+y^2sinx)dA, R=[-pi,pi]x[-pi,pi].
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]∫[-π,π] 1+x^2 siny + y^2 sinx dy dx
= ∫[-π,π] 2π dx
...
To evaluate the double integral using symmetry, we need to identify any symmetries in the given function and region of integration, and then simplify the integral accordingly.
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.
To evaluate the given double integral, we can use symmetry to simplify the calculation.
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
= 4π^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.