Question

Set up a double integral that gives the area of the surface on the graph of \(f\) over the region \(R\). \(f(x, y)=x^{3}-3 x y+y^{3}\) \(R\) : square with vertices (1,1),(-1,1),(-1,-1),(1,-1)

Short answer

The double integral set up for finding the surface area over the region \(R\) is \(\int_{-1}^{1} \int_{-1}^{1} \sqrt{1 + (3x^{2} - 3y)^{2} + (-3x + 3y^{2})^{2}} dx.dy\).

Step-by-step solution

Understanding the problem

This problem is composed of a function \(f(x, y)\) representing a surface in 3D space and a region \(R\) which is a square with given vertices. Our goal is to set up a double integral to compute the area of the surface of the function over the region \(R\).

Define the function and the region

The surface is defined by the function \(f(x, y) = x^{3} - 3xy + y^{3}\) and the region \(R\) is a square bounded by -1 ≤ x ≤ 1, -1 ≤ y ≤ 1.

Determine the differential area

The small area element in a two dimension integral is given by \(dx.dy\). Hence we multiply the function by this small area element to integrate over the entire region.

Setting up the double integral

We integrate the function \(f\) over the region \(R\) using two nested integrals accounting for both x and y directions. Consider the differentials in the order \(dx\) then \(dy\) (or vice versa), over the given limits. The double integral is set up as follows: \[ \int_{-1}^{1} \int_{-1}^{1} \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^{2} + \left(\frac{\partial f}{\partial y}\right)^{2}} dx.dy \]Where \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) are the partial derivatives of function \(f\) with respect to \(x\) and \(y\) respectively. These are computed as:\(\frac{\partial f}{\partial x} = 3x^{2} - 3y\) and \(\frac{\partial f}{\partial y} = -3x + 3y^{2}\)The final setting up of the double integral uses these results.

Simplifying the Double Integral Expression

The double integral hence becomes:\[ \int_{-1}^{1} \int_{-1}^{1} \sqrt{1 + (3x^{2} - 3y)^{2} + (-3x + 3y^{2})^{2}} dx.dy \]This is the set up of the double integral that gives the required area of the surface over the region \(R\). We have used derivatives and integral calculus in steps 1 to 5.