Problem 70
Question
Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.7 .\) Evaluate the improper integrals as iterated integrals. $$\int_{-1}^{1} \int_{-1 / \sqrt{1-x^{2}}}^{1 / \sqrt{1-x^{2}}}(2 y+1) d y d x$$
Step-by-Step Solution
Verified Answer
The value of the improper double integral is \(2\pi\).
1Step 1: Understand the Region of Integration
The integral is set over the region where \(-1 \leq x \leq 1\) and \(-1/\sqrt{1-x^2} \leq y \leq 1/\sqrt{1-x^2}\). This means for each fixed \(x\), \(y\) varies from \(-1/\sqrt{1-x^2}\) to \(1/\sqrt{1-x^2}\).
2Step 2: Solve the Inner Integral
Evaluate the inner integral with respect to \(y\):\[\int_{-1 / \sqrt{1-x^{2}}}^{1 / \sqrt{1-x^{2}}}(2y + 1) \, dy.\]Find the antiderivative of \(2y + 1\) which is \(y^2 + y\), and evaluate it at the bounds:\[[y^2 + y]_{-1 / \sqrt{1-x^{2}}}^{1 / \sqrt{1-x^{2}}}.\]This becomes:\[\left(\frac{1}{1-x^2} + \frac{1}{\sqrt{1-x^2}}\right) - \left(\frac{1}{1-x^2} - \frac{1}{\sqrt{1-x^2}}\right)\]Simplify to get:\[\frac{2}{\sqrt{1-x^2}}\].
3Step 3: Solve the Outer Integral
Now evaluate the outer integral with respect to \(x\):\[\int_{-1}^{1} \frac{2}{\sqrt{1-x^2}} \, dx.\]This integral represents the definite integral \(\int_{-1}^{1} \frac{2}{\sqrt{1-x^2}} \, dx\), which is equivalent to the integral of \(2\sin^{-1}(x)\) over its bounds. The corresponding indefinite integral is \(2\sin^{-1}(x)\) whose antiderivative, evaluated from \(-1\) to \(1\), accounts for:\[2\left(\sin^{-1}(1) - \sin^{-1}(-1)\right).\]Since \(\sin^{-1}(1) = \frac{\pi}{2}\) and \(\sin^{-1}(-1) = -\frac{\pi}{2}\), the calculation yields:\[2\left(\frac{\pi}{2} - (-\frac{\pi}{2})\right) = 2\pi.\]
4Step 4: Conclusion: Compute the Final Result
After evaluating both iterated integrals, the solution yields the value \(2\pi\). The integral converges to this value after taking into account the evaluation over the specified region of integration.
Key Concepts
Double IntegralsIterated IntegralsAntiderivatives
Double Integrals
Double integrals are an extension of definite integrals commonly used to compute the volume under a surface in a two-dimensional region. In practice, a double integral can be visualized as summing up infinitely many infinitesimally small rectangles, each contributing a tiny bit of volume, within a specified region. When performing a double integral, we take a function and integrate it over two variables, typically denoted as \(x\) and \(y\).
To conduct a double integral, you begin by considering the order of integration. One variable is integrated first, followed by the other. The region of integration is generally defined by boundaries for these variables, forming a closed area. In our example, the region is determined by the bounds for \(x\) and \(y\).
Double integrals can sometimes be improper, especially if the limits of integration tend to infinity or the function becomes undefined at some point in the region. However, you'd handle these improprieties like a single-variable improper integral, often using limits to define it properly.
To conduct a double integral, you begin by considering the order of integration. One variable is integrated first, followed by the other. The region of integration is generally defined by boundaries for these variables, forming a closed area. In our example, the region is determined by the bounds for \(x\) and \(y\).
Double integrals can sometimes be improper, especially if the limits of integration tend to infinity or the function becomes undefined at some point in the region. However, you'd handle these improprieties like a single-variable improper integral, often using limits to define it properly.
Iterated Integrals
Iterated integrals simplify the process of computing double integrals by breaking them down into two successive single integrals. You solve iterated integrals by integrating one variable at a time, known as the inner and outer integrals.
For our example, the iterated integral is processed by first handling the integral concerning \(y\). This inner integral focuses on the vertical slice of the region defined for each fixed \(x\). Once you compute this inner integral, you then use its result as the function for the outer integral. The outer integral is computed over the remaining variable, \(x\).
In our case, the antiderivative of the function \(2y + 1\) was evaluated over the specified limits of \(y\). The result was \(\frac{2}{\sqrt{1-x^2}}\), and this expression then serves as the function for the second integration with respect to \(x\).
For our example, the iterated integral is processed by first handling the integral concerning \(y\). This inner integral focuses on the vertical slice of the region defined for each fixed \(x\). Once you compute this inner integral, you then use its result as the function for the outer integral. The outer integral is computed over the remaining variable, \(x\).
In our case, the antiderivative of the function \(2y + 1\) was evaluated over the specified limits of \(y\). The result was \(\frac{2}{\sqrt{1-x^2}}\), and this expression then serves as the function for the second integration with respect to \(x\).
Antiderivatives
Antiderivatives, also known as indefinite integrals, are used extensively when solving integrals, including double and iterated ones. The process of finding an antiderivative involves determining a function whose derivative returns the given function.
When evaluating iterated integrals, you have to find antiderivatives for the inner and possibly the outer integrals. In our scenario, the antiderivative of the function \(2y + 1\) is required to solve the inner integral with respect to \(y\). The antiderivative here was found to be \(y^2 + y\). This antiderivative is crucial because it helps evaluate the definite integral over the \(y\) limits given by the problem.
When evaluating iterated integrals, you have to find antiderivatives for the inner and possibly the outer integrals. In our scenario, the antiderivative of the function \(2y + 1\) is required to solve the inner integral with respect to \(y\). The antiderivative here was found to be \(y^2 + y\). This antiderivative is crucial because it helps evaluate the definite integral over the \(y\) limits given by the problem.
- The expression \(y^2 + y\) is derived from finding an antiderivative of \(2y + 1\).
- This was evaluated over the limits \( -\frac{1}{\sqrt{1-x^2}} \) to \( \frac{1}{\sqrt{1-x^2}} \) to help solve the iterated integral.
Other exercises in this chapter
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