Question

Prove That The Equation F(X,Y,Z)=2z3−3(X2+Y2)Z Satisfies A Three-Dimensional Laplace Equation. (That Is: ∂X2∂2f+∂Y2∂2f+∂Z2∂2f=0 ).

Answer 1

After calculating the second partial derivatives of the function f(x,y,z) = 2z^3 - 3(x^2+y^2)z with respect to x, y, and z, it is shown that their sum equals zero, thus satisfying the three-dimensional Laplace equation.

Step-by-step explanation:

To prove that the equation f(x,y,z) = 2z^3 - 3(x^2+y^2)z satisfies the three-dimensional Laplace equation, we first need to take the partial second derivatives of f with respect to x, y, and z and then sum them.

First, the partial derivatives of f with respect to x and y are:

• ∂f/∂x = -6xz
• ∂f/∂y = -6yz

And then the second partial derivatives are:

• ∂²f/∂x² = -6z
• ∂²f/∂y² = -6z

The partial derivative of f with respect to z and its second derivative are:

• ∂f/∂z = 6z² - 3(x²+y²)
• ∂²f/∂z² = 12z

Combining these derivatives according to the Laplace equation, we get:

∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z² = (-6z) + (-6z) + (12z) = 0

Therefore, the equation f(x,y,z) = 2z^3 - 3(x^2+y^2)z does indeed satisfy the three-dimensional Laplace equation.

Answer 2

The function F(x, y, z) = 2z^3 - 3(x^2 + y^2)z satisfies the three-dimensional Laplace equation because the sum of its second partial derivatives with respect to x, y, and z equals zero.

Step-by-step explanation:

To prove that the function F(x, y, z) = 2z^3 - 3(x^2 + y^2)z satisfies the three-dimensional Laplace equation, we must show that the sum of the second partial derivatives with respect to x, y, and z equals zero. The Laplace equation in three dimensions is ∂^2f/∂x^2 + ∂^2f/∂y^2 + ∂^2f/∂z^2 = 0.

First, we find the second partial derivative of F with respect to x: ∂^2F/∂x^2 = -6z. Similarly, the second partial derivative with respect to y is ∂^2F/∂y^2 = -6z. The second partial derivative with respect to z is ∂^2F/∂z^2 = 12z.

Adding these derivatives up gives: -6z - 6z + 12z = 0. Hence, the function F(x, y, z) satisfies the three-dimensional Laplace equation.