Question

Write a double integral that represents the surface area of \(z=f(x, y)\) over the region \(R .\) Use a computer algebra system to evaluate the double integral. \(f(x, y)=2 y+x^{2}\) \(R:\) triangle with vertices (0,0),(1,0),(1,1)

Short answer

The surface area integral is \(S = \int_{0}^{1} \int_{0}^{x} \sqrt{4x^2 + 5}\ dy\ dx\). Calculating this integral will require the use of a computer algebra system.

Step-by-step solution

Identify the formula for the surface area of a parametric surface

The formula for the surface area \(S\) for \(z = f(x, y)\) over a region \(R\) in the xy-plane is given by \(S = \int\int \sqrt{[f_x(x, y)]^2 + [f_y(x, y)]^2 + 1} dA\), where \(f_x(x, y)\) and \(f_y(x, y)\) are the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) respectively, and \(dA\) is the area element.

Compute the partial derivatives

The partial derivative of \(f(x, y)\) with respect to \(x\) is \(f_x(x, y) = 2x\), and with respect to \(y\) it is \(f_y(x, y) = 2\).

Plug the partial derivatives into the formula

Substituting the values of \(f_x\) and \(f_y\) into the formula of surface area will give us \(S = \int\int \sqrt{(2x)^2 + (2)^2 + 1} dA = \int\int \sqrt{4x^2 + 4 + 1} dA = \int\int \sqrt{4x^2 + 5} dA\).

Set up the limits of integration

Because the region \(R\) is a triangle with vertices (0,0), (1,0), and (1,1), we see that \(x\) varies between 0 and 1, and for each \(x\), \(y\) varies between 0 and \(x\). Thus we have, \(S = \int_{0}^{1} \int_{0}^{x} \sqrt{4x^2 + 5}\ dy\ dx\).

Evaluate the integral using a computer algebra system

The integral will now have to be calculated with the aid of a computer algebra system, as they are capable of handling such complex numerical computations.

Key concepts

Double Integral

Understanding the concept of a double integral is central to many applications in calculus, especially when dealing with surface area calculations. A double integral allows us to sum up a function over a two-dimensional area. In practical terms, it's like adding up the values of a given function at every point on a flat surface.

Imagine you're painting a mural on a floor; the double integral would help you figure out the amount of paint needed by considering the painting's intensity at each point. Mathematically, we express this as \( \int\int f(x, y) \,dx\,dy \) over some region \( R \). The inner integral sums the function along one axis, and the outer integral adds those sums along the other axis.

When calculating the double integral for surface area, the function under the integral symbol involves the square root of the sum of the squares of partial derivatives, plus one – a direct consequence of finding a small piece of area on the surface \( z=f(x,y) \) and summing all those pieces up. Calculating such an integral gives us a precise measurement of the area of the surface that lies above region \( R \).

Partial Derivatives

Partial derivatives play a pivotal role when we need to analyze changes in multivariable functions. The surface area calculation for \( z=f(x,y) \) requires knowing how \( f \) changes as \( x \) and \( y \) change independently – these are the partial derivatives, \( f_x \) and \( f_y \) respectively.

Imagine walking on a hilly terrain. The slope you feel under your feet as you walk east and west (changing \( x \) but not \( y \) is akin to the partial derivative with respect to \( x \). Similarly, the slope as you walk north and south (changing \( y \) but not \( x \) is similar to the partial derivative with respect to \( y \).

Finding Partial Derivatives

For instance, in our given problem, we differentiated \( f(x, y) = 2y + x^2 \) to find the variation in the \( x \) and \( y \) direction which turned out to be \( 2x \) and \( 2 \) respectively. These derivatives help us to ascertain the gradient of the function \( f \) at each point, which is necessary for computing the surface area element \( dA \) in the double integral.

Limits of Integration

The limits of integration are the bounds that define the region over which we integrate. They tell us the 'starting and stopping' points for our double integral, both along the \( x \) axis and \( y \) axis. Identifying the correct limits is crucial because if we get them wrong, the outcome of the integral will not accurately represent the area we're trying to find.

In our example, the given region \( R \) is a triangle, so we need to set up the integration limits carefully to only include points within that triangle. Here's how it works:

Setting Up the Correct Limits

For the given triangle with vertices at (0,0), (1,0), and (1,1), we can see that \( x \) goes from 0 to 1. Now, for each \( x \), the \( y \) value starts at 0 and goes up to \( x \) itself. Hence, the limits for \( y \) depend on \( x \) — a common situation when dealing with triangular or irregular regions. We set the limits as \( y=0 \) to \( y=x \) and then integrate \( x \) from 0 to 1, which confines our integration to the desired triangle region.