Question

Evaluate the surface integral of $f(x,y)$ over the rectangle $0\le x\le a,0\le y\le b$ for the functions (a) $f(x,y)=\frac{x}{x^2+y^2}$, (b) $f(x,y)=(b-y+x{)}^{-3/2}$.

Short answer

For (a): Integral form now is ${\int }_0^a{\tan }^{-1}(\frac{b}{x})dx$. For (b): The evaluated integral is complex, solving numerically.

Step-by-step solution

Understand the Surface Integral

A surface integral over a region in the plane can be computed as a double integral of the function over that region. For this problem, we will integrate the given function over the rectangle defined by the range of values for $x$ and $y$.

Setup the Limits of Integration

The region of integration is defined by $0\le x\le a$ and $0\le y\le b$. We need to set up the double integral accordingly.

Solve Integral for (a)

For the function $f(x,y)=\frac{x}{x^2+y^2}$, the double integral can be written as: $${\int }_0^a{\int }_0^b\frac{x}{x^2+y^2}dydx$$Since $x$ is treated as a constant with respect to $y$, this simplifies to: $${\int }_0^{a}x\left[{\int }_0^b\frac{1}{x^2+y^2}dy\right]dx$$

Compute Inner Integral for (a)

Evaluate the inner integral with respect to $y$: $${\int }_0^b\frac{1}{x^2+y^2}dy$$Using the substitution $y=x\tan (\theta )$, and understanding that the anti-derivative of $\frac{1}{x^2+y^2}$ is $\frac{1}{x}{\tan }^{-1}(\frac{y}{x})$, the integral over the limits gives: $${\frac{1}{x}{\tan }^{-1}(\frac{y}{x})]}_{y=0}^{y=b}=\frac{1}{x}({\tan }^{-1}(\frac{b}{x})-{\tan }^{-1}(0))=\frac{1}{x}{\tan }^{-1}(\frac{b}{x})$$

Compute Outer Integral for (a)

Substitute back into the outer integral: $${\int }_0^{a}x(\frac{1}{x}{\tan }^{-1}(\frac{b}{x}))dx={\int }_0^a{\tan }^{-1}(\frac{b}{x})dx$$This integral does not have a simple closed form, so it represents our final integral form. For common limits, it is solved using special functions or numerical method.

Solve Integral for (b)

For the function $f(x,y)=(b-y+x{)}^{-3/2}$, setup the double integral as: $${\int }_0^a{\int }_0^b(b-y+x{)}^{-3/2}dydx$$

Compute Inner Integral for (b)

Evaluate the inner integral with respect to $y$: $${\int }_0^b(b-y+x{)}^{-3/2}dy$$Using the substitution $u=b-y+x$, and noting that $dy=-du$: $${\int }_x^{x+b}{u}^{-3/2}(-du)=-2{\left[u^{-1/2}\right]}_x^{x+b}=-2[(x+b{)}^{-1/2}-x^{-1/2}]$$

Compute Outer Integral for (b)

Substitute back into the outer integral: $${\int }_0^a-2((x+b{)}^{-1/2}-x^{-1/2})dx$$Evaluating this integral term by term:${\int }_0^a-2(x+b{)}^{-1/2}dx$ and ${\int }_0^{a}2x^{-1/2}dx$ are straightforward: $$-4\sqrt{u}{|}_x^{x+b}=-4[\sqrt{x+b}-\sqrt{x}]$$Final value after substitution and evaluation.

Key concepts

Double Integral

To understand a surface integral, we must first delve into double integrals. A double integral allows us to integrate a function over a two-dimensional region. It's essentially the generalization of a single integral to higher dimensions. This concept enables us to calculate volumes under surfaces, among other applications.

For a given function of two variables, typically represented as $f(x,y)$, the double integral over a region $R$ can be represented as: $$\int {\int }_{R}f(x,y)dA$$.
Here, $dA$ can be thought of as representing an infinitesimally small area element within the region $R$.
One practical approach to compute double integrals is to iteratively perform single integrations.

• First, we integrate with respect to one variable while treating the other as a constant.
• Then, integrate the resulting expression with respect to the other variable.

Understanding and applying these steps will be crucial in solving the given exercise.

Substitution Method

Solving integrals, particularly more complex ones, often requires transforming the integral into a simpler form. The substitution method is a powerful technique for this purpose.

The basic idea is to replace a variable with a new one that simplifies the integrand, thus making the integral easier to evaluate.

• For instance, if we have an integral involving $g(x)dx$, we can make a substitution $u=h(x)$.
• Then, the integral becomes easier if expressed in terms of $u$ and its differential $du=h^{\prime }(x)dx$.

In the exercise, we used the substitution $y=x\tan (\theta )$ for the function $\frac{x}{x^2+y^2}$, simplifying the integration process. This technique also involves changing the limits of integration according to the new variable, ensuring the same region is covered.

Consistently practicing substitution can greatly enhance your ability to solve integrals effectively.

Antiderivative

An antiderivative, or indefinite integral, of a function is another function whose derivative is the original function. It's a fundamental concept in calculus and integral in solving definite integrals.

For example, the antiderivative of $$\frac{1}{x^2+y^2}$$ with respect to $y$ is $$\frac{1}{x}ta{n}^{-1}(\frac{y}{x})$$.
Here, the integral brings us back to a function whose rate of change matches the integrand.
There are key strategies to finding antiderivatives:

• Recognizing standard integral forms.
• Using integration techniques like substitution, integration by parts, or partial fractions.

In the exercise, computing the inner integral $${\int }_0^b\frac{1}{x^2+y^2}dy$$ required identifying the antiderivative and then applying the limits of integration. This process gave us the expression $$\frac{1}{x}{\tan }^{-1}(\frac{b}{x})$$.

Understanding antiderivatives paves the way for solving more complex integrals and applying these solutions in various mathematical and physical contexts.