Question

Sketch the region of integration and evaluate the following integrals as they are written. $$\int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} d x d y$$

Short answer

Answer: The value of the given double integral is \(e - 1\).

Step-by-step solution

Sketch the Region of Integration

The limits of integration for \(x\) in the inner integral are from \(0\) to \(\cos y\). The limits for \(y\) in the outer integral are from \(0\) to \(\pi / 2\). Let's first find the limits for \(x\), which are \(0\) and \(\cos y\). At \(y = 0\), we have \(x = \cos(0) = 1\), and at \(y = \frac{\pi}{2}\), we have \(x = \cos(\frac{\pi}{2}) = 0\). So, the region to be integrated consists of points \((x,y)\) where \(0\leq x \leq \cos y\) and \(0 \leq y \leq \frac{\pi}{2}\). We can visualize the region as follows: Start at the origin \((0, 0)\). The lower limit of the \(y\) integral is a horizontal line at \(y=0\). The upper limit of the \(y\) integral is a horizontal line at \(y = \frac{\pi}{2}\). Now, for a given \(y\), the function \(x = \cos y\) is the upper limit of the \(x\) integral, and \(x = 0\) is the lower limit. So the region lies between the curve \(x = \cos y\), the \(y\)-axis and the lines \(y=0\) and \(y=\frac{\pi}{2}\).

Evaluation of the Inner Integral

Now we will evaluate the integral inside the square brackets. In this case, the inner integral is with respect to \(x\), so we will integrate the function \(e^{\sin y}\) with respect to \(x\): $$\int_{0}^{\cos y} e^{\sin y} dx$$ Since \(e^{\sin y}\) is a constant with respect to \(x\) (it only depends on \(y\)), this integral becomes: $$e^{\sin y} \int_{0}^{\cos y} dx = e^{\sin y} [x]_{0}^{\cos y} = e^{\sin y}(\cos y) - e^{\sin y}(0) = e^{\sin y}\cos y$$

Evaluation of the Outer Integral

Now our integral is simplified to a single integral with respect to \(y\): $$\int_{0}^{\pi / 2} e^{\sin y}\cos y dy$$ To evaluate this integral, we can use 'Integration by Substitution'. Let's make the substitution \(u = \sin y\). Then, the differential \(du = \cos y dy\). Now, when \(y = 0\), \(u = \sin(0) = 0\) and when \(y = \pi / 2\), \(u = \sin(\frac{\pi}{2}) = 1\). Our integral becomes: $$\int_{0}^{1} e^u du$$ Now, we can integrate this simple integral: $$\int_{0}^{1} e^u du = [e^u]_{0}^{1} = e^1 - e^0 = e - 1$$ Thus, the value of the given double integral is: $$\int_{0}^{\pi / 2} \int_{0}^{\cos y} e^{\sin y} dx dy = e - 1$$

Key concepts

Region of Integration

In the world of double integrals, the region of integration is crucial for determining the area over which you're performing the integration. In this problem, we are dealing with a specific set of bounds for our variables, making it essential to visualize where the integration happens. Let's break it down:

• The inner integral, with respect to \(x\), is limited between \(0\) and \(\cos y\). This means for each fixed \(y\), the value of \(x\) ranges from \(0\) to \(\cos y\).
• The outer integral, set for \(y\), allows \(y\) to vary from \(0\) to \(\pi/2\).

Visualizing this, we can sketch a region on the coordinate plane. This area is confined by the \(y\)-axis, the curve defined by \(x = \cos y\), the horizontal line \(y = 0\), and the line \(y = \frac{\pi}{2}\). Essentially, it is a region bounded by a declining cosine curve and horizontal limits on a \([0, 1]\) interval for \(x\) when \(y\) varies over the specified range.

Integration by Substitution

The integration by substitution technique is a powerful tool that simplifies the process of integration by changing variables. In this exercise, it is used to solve the outer integral, which involves a trigonometric function that complicates direct integration. Here’s how it was applied:

• The integral \(\int_{0}^{\pi / 2} e^{\sin y} \cos y \, dy\) initially seems challenging because of the product of \(\cos y\) and \(e^{\sin y}\).
• By setting \(u = \sin y\), the derivative becomes \(du = \cos y \, dy\), allowing the integral to be rewritten with respect to \(u\).
• The bounds also change with this substitution. When \(y = 0\), \(u = \sin 0 = 0\), and when \(y = \frac{\pi}{2}\), \(u = \sin \frac{\pi}{2} = 1\).
• This transforms the integral into a simpler exponential form: \(\int_{0}^{1} e^u \, du\).

This substitution simplifies integration to an easily solvable exponential form, showcasing why substitution is such a favored technique in calculus.

Calculus

Calculus, the mathematical study of change, is essential for solving integrals like the ones in this problem. It involves several methods and concepts crucial to understanding the world of continuous mathematics:

• **Differentiation** focuses on understanding the rates at which things change. While it might not be directly used in this integral, it informs the understanding of substitution and variable changes.
• **Integration**, its counterpart, deals with accumulation. In particular, double integrals are used to compute volumes under surfaces or other higher-dimensional accumulations.
• **Integral bounds** help in understanding the limits over which change is observed, as seen with our region of integration. This sets the total area or volume under consideration.

In essence, calculus provides tools for solving complex real-world problems by breaking down and analyzing change through differentiation and accumulation through integration. It gives us the means to transform difficult integrals into manageable calculations, just as we did in this exercise.