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. $$ \begin{array}{l} f(x, y)=\frac{2}{3} x^{3 / 2}+\cos x \\ R=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq 1\\} \end{array} $$

Short answer

The double integral is \(\int_0^1 \int_0^1 \sqrt{1+(\sqrt{x}-\sin(x))^2}\, dx\, dy\). To evaluate it to get the surface area, a computer algebra system is required.

Step-by-step solution

Find the Surface Differential

To calculate the surface area, we first find the surface differential, which is given by \(dS=\sqrt{1+(\nabla f)^2}dx\, dy\), where \(\nabla f\) is the magnitude of the gradient of \(f\). So, first derive \(f\) partially w.r.t. \(x\) and \(y\): \[\frac{\partial f}{\partial x}=\frac{2}{3}\cdot\frac{3}{2}\sqrt{x} - \sin(x) = \sqrt{x} - \sin(x)\] and \[\frac{\partial f}{\partial y} = 0\] Now, calculate the magnitude of the gradient: \[\nabla f=\sqrt{(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2} = \sqrt{[f_x(x,y)]^2 + 0^2} = |\sqrt{x} - \sin(x)|\]. Insert \(\nabla f\) into the equation for the surface differential to obtain: \(dS=\sqrt{1+( \sqrt{x}- \sin(x) )^2 }dx\, dy\).

Setup the Double Integral

Now we can set up the double integral which will give us the area of the surface. From the limits given in the region \(R\), we know that \(x\) and \(y\) are both limited from 0 to 1. Our double integral will then be: \[Int_s dS = \int_0^1 \int_0^1 \sqrt{1+(\sqrt{x}-\sin(x))^2}dx\, dy\].

Compute the Double Integral Using CAS

At this point we would use Computer Algebra System(CAS) to perform the integration since it's fairly complex with non-basic integral. The integral may not have a simple closed form and might be represented as a numeric approximation.

Key concepts

Surface Differential

Understanding the surface differential is crucial when it comes to finding the surface area of a function. Imagine draping a piece of fabric over a three-dimensional shape; the surface differential, represented as \(dS\), measures a tiny piece of that fabric. Mathematically, \(dS\) is used to capture the element of the surface area of the graph of a function, such as \(z=f(x, y)\).

To find the surface differential, we take into account the gradient of the function, which we'll discuss in more detail later. In the case of our double integral problem, the formula for \(dS\) is \(dS=\sqrt{1+(abla f)^2}dx\, dy\). Here, \(abla f\) is the magnitude of the gradient of the function \(f\), which depends on the partial derivatives with respect to \(x\) and \(y\). When the function is only changing with one variable, as with \(f(x, y)\) in our exercise, where the partial derivative with respect to \(y\) is zero, the surface differential simplifies to account for changes in the single variable.

Gradient of a Function

The gradient of a function points in the direction of the steepest ascent. If you're hiking, following the gradient would lead you directly uphill. It is a vector that we denote by \(abla f\), and it is composed of the partial derivatives of a function with respect to its variables.

In our context, we're looking at how function \(f(x, y)\) changes as we move in the \(x\)- and \(y\)-directions. The gradient is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). In the textbook exercise, since \(\frac{\partial f}{\partial y} = 0\), our gradient simplifies to just \(\frac{\partial f}{\partial x}\).

The magnitude of the gradient, \(|abla f|\), is part of the formula for the surface differential and greatly influences the calculated area. When we say \(\sqrt{1+(abla f)^2}\), we are accounting for not just the flat projection of the area onto the xy-plane, but also how 'steep' the surface is at any point.

Partial Derivatives

Partial derivatives are a way to see how a function changes in one direction, holding the other variables constant. Imagine you're standing on a hillside, the rate at which the hill rises or falls as you walk east is like a partial derivative in the \(x\)-direction.

In our exercise, we found two partial derivatives: \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). These derivatives tell us about the slope of the function \(f\) in the directions of \(x\) and \(y\), respectively. However, because our function doesn't change with \(y\), \(\frac{\partial f}{\partial y} = 0\), indicating no slope in the \(y\)-direction. The non-zero partial derivative, \(\frac{\partial f}{\partial x} = \sqrt{x} - \sin(x)\), then becomes vital in calculating the surface differential since it's the only factor affecting the gradient in our case.

Computer Algebra System

A Computer Algebra System (CAS) is a powerful tool that can manipulate and solve complex mathematical expressions symbolically. Think of it as a highly advanced calculator that can perform integrations, solve equations, and even plot graphs.

In the final step of our textbook exercise, we turn to a CAS to tackle the double integral. This is because the integration involved is not straightforward and would be extremely tedious, if not impossible, to perform by hand. The beauty of a CAS lies in its ability to handle the nitty-gritty computational work, giving us more time to understand the concepts at play. When using a CAS, we input our problem and it provides us with an exact or numerical answer. In some cases, such as when we expect the integral to not have a simple closed form, a CAS may offer a numerical approximation, which is perfectly acceptable for understanding the general behavior or solution.