Problem 18 - University Calculus: Early Transcendentals

Question

Evaluate the double integral over the given region \(R\). $$\iint_{R} y \sin (x+y) d A, \quad R: \quad-\pi \leq x \leq 0, \quad 0 \leq y \leq \pi$$

Step-by-Step Solution

Answer

The double integral evaluates to 4.

Step 1: Set Up the Double Integral

The region of integration is defined as \( R: -\pi \leq x \leq 0 \) and \( 0 \leq y \leq \pi \), which represents a rectangle in the \(xy\)-plane. We express the double integral using these limits:\[\iint_{R} y \sin (x+y) \, dA = \int_{0}^{\pi} \int_{-\pi}^{0} y \sin(x+y) \, dx \, dy\]This sets up the double integral with the inner integral with respect to \(x\) from \(-\pi\) to \(0\) and the outer integral with respect to \(y\) from \(0\) to \(\pi\).

Step 2: Integrate with Respect to \(x\)

The next step is to evaluate the inner integral, which is:\[\int_{-\pi}^{0} y \sin(x+y) \, dx\]Use the substitution \( u = x + y \), \( du = dx \). When \( x = -\pi \), \( u = -\pi + y \) and when \( x = 0 \), \( u = y \). Therefore, the limits of integration for \(u\) are from \(-\pi + y\) to \(y\), and the integral becomes:\[y \int_{-\pi + y}^{y} \sin(u) \, du\]Evaluating the integral gives:\[y \left[ -\cos(u) \right]_{-\pi+y}^{y} = y \left[ -\cos(y) + \cos(-\pi + y) \right]\]Using the property \( \cos(-\theta) = \cos(\theta) \), we have:\[y (2\cos(y))\]

Step 3: Integrate with Respect to \(y\)

Now, we integrate the result from the previous step with respect to \(y\):\[\int_{0}^{\pi} y (2\cos(y)) \, dy\]This simplifies to:\[2 \int_{0}^{\pi} y \cos(y) \, dy\]We can use integration by parts, letting \( u = y \) and \( dv = \cos(y) \, dy \). Then \( du = dy \) and \( v = \sin(y) \). Integrating by parts:\[u\, dv = uv - \int v \, du = y\sin(y)\bigg|_{0}^{\pi} - \int_{0}^{\pi} \sin(y) \, dy\]The term \( y \sin(y) \big|_{0}^{\pi} = 0 \) because \( y \sin(y) \big|_{\pi} = 0 \) and \( y \sin(y) \big|_{0} = 0 \). Hence, the integral becomes:\[-\left[-\cos(y)\right]_{0}^{\pi} = 1 + 1 = 2\]Thus, the integral evaluates to \( 2(2) = 4 \).

Step 4: Final Answer

The double integral, therefore, evaluates to 4, as calculated. So the final answer for the double integral over the region \( R \) is 4.

Key Concepts

Integration by Parts

The method of integration by parts is a vital tool for evaluating integrals, especially when they involve products of functions, such as the one we encounter in our exercise. The formula behind this method is derived from the product rule of differentiation and is given as:\[ \int u \, dv = uv - \int v \, du \]Here, you get to choose which part of the integrand will be \(u\) and which will be \(dv\). In our exercise, we used integration by parts to evaluate the integral of \( y\cos(y) \). We set:

• \( u = y \), \( du = dy \)
• \( dv = \cos(y) \, dy \), which gives \( v = \sin(y) \)

After applying integration by parts, we find that the boundary terms \( y \sin(y) \big|_0^\pi \) cancel out to zero because they equal zero at both boundaries. This left us with a simpler integral to evaluate. Integration by parts often simplifies an otherwise complex integral, making it easier to solve.

Region of Integration

Understanding the region of integration is important for correctly setting up the limits of a double integral. The given region in the exercise is defined in the \(xy\)-plane by the bounds \(-\pi \leq x \leq 0\) and \(0 \leq y \leq \pi\). This delimits a rectangular region.When expressing a double integral, the region of integration determines the order of integration and the corresponding integration limits. In our case, the inner integral was taken over \(x\) from \(-\pi\) to \(0\) and for each fixed value of \(x\), \(y\) was integrated from \(0\) to \(\pi\). Setting these bounds properly ensures that each point in the desired region is covered exactly once. This understanding helps prevent mistakes when solving double integrals.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving trigonometric functions. In our step-by-step solution, the substitution \( u = x + y \) was used to simplify the integral \( \int y \sin(x+y) \, dx \). This transforms the integral into terms involving \(u\), which is easier to solve.When doing such substitutions:

• Change the integration variable and limits to match those of \(u\).
• Express \(dx\) in terms of \(du\). In this scenario, as \( u = x + y \), \( du = dx \), making it a straightforward change of variables.
In the substitution, the limits of integration were transformed from \(x\) being \([-\pi, 0]\) to \(u\) being \([-\pi + y, y]\), simplifying the process of directly dealing with trigonometric integrals. Substitution, especially in combination with trigonometric identities, can significantly simplify the problem when dealing with complex integral expressions.