Question

Evaluate (a) \(\int_{S}\left(2 x+y^{2}\right) d(x, y) ; \quad S=\left\\{(x, y) \mid 0 \leq x \leq 9-y^{2},-3 \leq y \leq 3\right\\}\) (b) \(\int_{S} 2 x y d(x, y) ; \quad S\) is bounded by \(y=x^{2}\) and \(x=y^{2}\) (c) \(\int_{S} e^{x} \frac{\sin y}{y} d(x, y) ; \quad S=\\{(x, y) \mid \log y \leq x \leq \log 2 y, \pi / 2 \leq y \leq \pi\\}\)

Short answer

#Answer#: (a) The value of the integral is \(\frac{486}{5}\). (b) The value of the integral is \(\frac{1}{12}\). (c) The value of the integral is \(1\).

Step-by-step solution

Identify the bounds of integration

The region \(S=\{(x, y)\mid 0 \leq x \leq 9-y^2, -3 \leq y \leq 3\}\) can be described as integrating over \(y\) from \(-3\) to \(3\) and integrating over \(x\) from \(0\) to \(9-y^2\). We will have the following iterated integral: $$\int_{-3}^{3}\int_{0}^{9-y^2}(2x+y^2) dx dy$$

Perform the inner integral

Integrating with respect to \(x\) gives: $$\int_{-3}^{3} \left[x^2+xy^2\right]_{0}^{9-y^2} dy$$

Simplify and perform the outer integral

Plugging in the bounds for \(x\), we get: $$\int_{-3}^{3}(81-18y^2+y^4) dy$$ Integrating with respect to \(y\) gives: $$\left[\frac{81y}{2}-6y^3+\frac{y^5}{5}\right]_{-3}^3$$

Compute the final answer

Plugging in the bounds for \(y\), we have: $$\left[\frac{81(3)}{2}-6(3)^3+\frac{(3)^5}{5}\right]-\left[\frac{81(-3)}{2}-6(-3)^3+\frac{(-3)^5}{5}\right]=\frac{486}{5}$$ Thus, the value of the integral is \(\frac{486}{5}\). (b) Evaluate the integral over the region bounded by \(y=x^2\) and \(x=y^2\).

Identify the bounds of integration

First, notice that \(y=x^2\) and \(x=y^2\) intersect when \(x=y^2 \Rightarrow y^2=x^2 \Rightarrow y=\pm x\). To find the bounds of integration, consider integrating over \(y\) from \(x^2\) to \(\sqrt{x}\), and integrating over \(x\) from \(0\) to \(1\) (since the intersection occurs at \((0, 0)\) and \((1, 1)\)). The iterated integral is then: $$\int_{0}^{1}\int_{x^2}^{\sqrt{x}} 2xy dy dx$$

Perform the inner integral

Integrating with respect to \(y\): $$\int_{0}^{1}\left[x\cdot \frac{y^2}{2}\right]_{x^2}^{\sqrt{x}} dx$$

Simplify and perform the outer integral

Plugging in the bounds for \(y\), we get: $$\int_{0}^{1}\left[\frac{1}{2}x^{-1/2}-x^3\right] dx$$ Integrating with respect to \(x\): $$\left[\frac{1}{3}x^{1/2}-\frac{1}{4}x^4\right]_0^1 = \frac{1}{3}-\frac{1}{4}$ The value of the integral is \(\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\). (c) Evaluate the integral over the given region.

Identify the bounds of integration

The region \(S=\{(x, y)\mid \log y \leq x \leq \log 2y, \frac{\pi}{2} \leq y \leq \pi\}\) can be described as integrating over \(y\) from \(\frac{\pi}{2}\) to \(\pi\), and integrating over \(x\) from \(\log y\) to \(\log 2y\). The iterated integral is then: $$\int_{\pi/2}^{\pi}\int_{\log y}^{\log 2y} e^x\frac{\sin y}{y} dx dy$$

Perform the inner integral

Integrating with respect to \(x\): $$\int_{\pi/2}^{\pi}\left[e^x\right]_{\log y}^{\log 2y} \frac{\sin y}{y} dy$$

Simplify and perform the outer integral

Plugging in the bounds for \(x\), we get: $$\int_{\pi/2}^{\pi}\left(2y- y\right) \frac{\sin y}{y} dy = \int_{\pi/2}^{\pi} \sin(y) dy $$ Integrating with respect to \(y\): $$\left[-\cos y\right]_{\pi/2}^{\pi} = 1 - 0 = 1$$ The value of the integral is \(1\).

Key concepts

Iterated Integral

An iterated integral is a mathematical tool used to calculate the volume under a surface on a given region in a multi-dimensional space. In layman's terms, when you are trying to find the area under a curve in two dimensions, you use a regular integral. But when the problem involves a surface (which is like a stretched sheet over a region) in three dimensions, we often resolve to iterated integrals. They are applied sequently, one after the other, typically starting with the innermost integral. For example, for the given function over region S, we evaluate the iterated integral as shown below:

1. For a region S, define the order of integration based on the limits of the variables involved.
2. Compute the inner integral first, usually by treating the other variables as constants.
3. After finding the result of the inner integral, you then solve the outer integral, which will give you the final answer.

When solving part (a) of the exercise, you begin by tackling the inner integral over x while treating y as a constant. Once the inner integral is computed, you integrate the resulting expression over y to find the final answer.

Bounds of Integration

The bounds of integration, or limits of integration, are values that specify the start and end points for which you evaluate an integral. In the context of double integrals, we have two variables, which means we need to establish bounds for both. The bounds are dictated by the region S over which we're integrating the function.

When you integrate, you essentially add up infinitesimally small quantities across a given range – the bounds of your integration set this range. Correctly identifying the bounds for each variable is crucial, as it ensures you're considering the entire region and only the region you're interested in.

The region S is often described by inequalities involving the variables x and y, as seen in the exercise solutions, and can sometimes be depicted visually as the area enclosed by curves on a graph. Determining where these curves intersect can help in establishing the appropriate bounds.

Real Analysis

Real analysis is a branch of mathematics dealing with real numbers and the analytic properties of real functions and sequences. It encompasses theories such as convergence, limits, integration, and differentiation. In the context of our problem, real analysis provides the foundation for understanding concepts like multiple integrals and the reasoning behind the methods used to solve them.

As part of real analysis, we often explore different types of convergence, such as pointwise and uniform convergence, and how they apply to swapping the order of limits and integrals. Another important aspect of real analysis that's relevant to our exercise is the study of improper integrals, where one or both limits of integration extend to infinity or the function has singularities.

Multiple Integral

A multiple integral extends the concept of a single integral to functions of multiple variables, enabling us to evaluate volume under a surface in higher dimensions. While double integrals are specific to functions with two independent variables, triple integrals deal with functions of three variables, and so on.

When it comes to evaluating a multiple integral, the process generally involves taking an iterated approach: first solving an integral with respect to one variable while treating the others as constants, and then proceeding to the next variable in a cascading fashion until all integrals have been resolved.

In the given exercise, we see examples of double integrals, as we integrate a function over a two-dimensional region. Each problem provides a different region of integration, which requires setting up the correct bounds of integration and understanding how to integrate over a planar area, reflecting the essence of multiple integrals in real-world applications.