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Chapter 13: Problem 94

Many improper double integrals may be handled using the techniques for improper integrals in one variable (Section 7.8 ). For example, under suitable conditions on \(f\), \(\int_{a}^{\infty} \int_{g(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x\). Use or extend the one-variable methods for improper integrals to evaluate the following integrals. $$\int_{1}^{\infty} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x$$

Short Answer

Expert verified
To evaluate the given double improper integral, we first converted it into a proper integral by introducing a limit. Then, we evaluated the inner integral with respect to y, being left with a single integral multiplied by the limit. Lastly, we integrated with respect to x and evaluated the limit to find the value of the double integral as 1/2.

Step by step solution

01

Identify the improper integral

The given double integral is: $$\int_{1}^{\infty} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x$$ Notice that the outer integral has an upper limit of \(\infty\), which is an indication of an improper integral.
02

Convert the improper integral to proper integral form using limits

To convert the double integral into a proper integral, we use a limit as follows: $$\lim_{b \rightarrow \infty} \int_{1}^{b} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x$$
03

Evaluate the inner integral

First, we will evaluate the inner integral with respect to \(y\): $$\int_{0}^{1 / x^{2}} \frac{2 y}{x} d y$$ As the integrand has no \(x\) terms, the integration is straightforward: $$\frac{2}{x}\int_{0}^{1 / x^{2}} y d y$$ Now, integrate with respect to \(y\) and evaluate the limits: $$\frac{2}{x}\left[\frac{1}{2}y^2\right]_{0}^{1 / x^{2}} = \frac{2}{x}\cdot \frac{1}{2}\cdot\left(\frac{1}{x^2}\right)^2 - 0 = \frac{1}{x^3}$$
04

Evaluate the outer integral with the limit

We now have a single integral to evaluate along with the limit: $$\lim_{b \rightarrow \infty} \int_{1}^{b} \frac{1}{x^3} d x$$ Integrate with respect to \(x\) and evaluate the limit: $$\lim_{b \rightarrow \infty} \left[-\frac{1}{2x^2}\right]_{1}^{b} = \lim_{b\rightarrow \infty}\left(-\frac{1}{2b^2} + \frac{1}{2}\right)$$ As \(b \rightarrow \infty\), the term \(\frac{1}{2b^2}\) approaches 0: $$-\frac{1}{2\cdot\infty^2} + \frac{1}{2} = 0+\frac{1}{2} = \frac{1}{2}$$ So, the value of the double integral is: $$\int_{1}^{\infty} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x = \frac{1}{2}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integrals
Double integrals are a fascinating aspect of calculus, particularly useful for finding volumes under surfaces defined by functions of two variables, like \(f(x, y)\). In the given exercise, the double integral \\[\int_{1}^{\infty} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x\]\helps us understand how functions behave over an infinite region. This involves:
  • The outer integral, which sums the results of the inner integrals as the variable (xgoes from 1 to infinity.
  • The inner integral, which evaluates the function from a lower boundary of zero to an upper boundary that changes with x,specifically \(1/x^2\).
Don't worry, even though there's an infinity involved, we manage it with limits, making it a solvable problem!
Integration Techniques
Integration techniques are the methods used to solve integrals, which can sometimes be tricky, especially with functions that involve infinity, like our improper integrals. Here are some key steps:
  • First, recognize if the problem is related to improper integrals. This often happens when infinity appears as one or more of the integration limits.
  • Next, convert the integral into a form that makes it easier to handle. This typically involves using a limit to replace the infinite bounds in the integral.
  • Afterward, evaluate the inner integral to simplify the problem. In our case, focusing on \(y\) first helps us see that the inner integral results in \(1/x^3\).
Mastering these techniques allows for flexibility and creativity in solving complex mathematical problems.
Limits in Calculus
Understanding limits is crucial for tackling improper integrals like the one in this exercise. Limits deal with the behavior of a function as it approaches a certain point鈥攆or example, as xapproaches infinity.
When evaluating improper integrals:
  • First, express the infinite process using a limit, such as \(\lim_{b \to \infty} \int_{1}^{b} \), to transform the problem.
  • Continue by solving the integral as if the limit wasn't initially involved, integrating respect to x,and substitute back to find the final value.
  • Finally, apply the limit to conclude the evaluation. In our solution, as \b\ tends to infinity, \(1/2b^2\) becomes negligible, leading to the result \1/2.This simplification through limits makes it manageable!
Limits transform seemingly endless problems into concise and calculable solutions.

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Most popular questions from this chapter

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid rectangular box has sides of length \(a, b,\) and \(c .\) Where is the center of mass relative to the faces of the box?

Which bowl holds more water if it is filled to a depth of 4 units? \(\cdot\) The paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4\) \(\cdot\) The cone \(z=\sqrt{x^{2}+y^{2}},\) for \(0 \leq z \leq 4\) \(\cdot\) The hyperboloid \(z=\sqrt{1+x^{2}+y^{2}},\) for \(1 \leq z \leq 5\)

Intersecting spheres One sphere is centered at the origin and has a radius of \(R\). Another sphere is centered at \((0,0, r)\) and has a radius of \(r,\) where \(r>R / 2 .\) What is the volume of the region common to the two spheres?

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Water in a gas tank Before a gasoline-powered engine is started, water must be drained from the bottom of the fuel tank. Suppose the tank is a right circular cylinder on its side with a length of \(2 \mathrm{ft}\) and a radius of 1 ft. If the water level is 6 in above the lowest part of the tank, determine how much water must be drained from the tank.
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