Problem 51 - Calculus Early Transcendentals

Question

Let \(R=\\{(x, y): 0 \leq x \leq \pi\) \(0 \leq y \leq a\\} .\) For what values of \(a,\) with \(0 \leq a \leq \pi,\) is \(\iint_{R} \sin (x+y) d A\) equal to \(1 ?\)

Step-by-Step Solution

Answer

Question: Find the values of \(a\) such that the double integral of \(\sin(x+y)\) over the region R is equal to 1. The region \(R=\{(x, y): 0 \leq x \leq \pi\) , \(0 \leq y \leq a\}\). Answer: \(a = \frac{3\pi}{2}\).

Step 1: Set up the double integral

We are given the region \(R\), so we can set up the double integral as follows: \(\iint_{R} \sin(x+y) dA.\) In this case, the differential area \(dA = dxdy\), so the double integral becomes: \(\int_{0}^{\pi} \int_{0}^{a} \sin(x+y)\, dy\, dx.\)

Step 2: Evaluate the inner integral

Let's first evaluate the inner integral (over \(y\)): \(\int_{0}^{a} \sin(x+y)\, dy.\) To evaluate this integral, we'll make a substitution. Let \(u = x + y\). Then, \(du = dy\), and when \(y = 0\), \(u = x\), and when \(y = a\), \(u = x + a\). Now, the integral becomes: \(\int_{x}^{x+a} \sin(u)\, du = [-\cos(u)]_{x}^{x+a} = -\cos(x+a) + \cos{x}\).

Step 3: Evaluate the outer integral

Now, we need to plug the result from the inner integral back into the outer integral and evaluate it: \(\int_{0}^{\pi}[-\cos(x+a) + \cos{x}]\, dx.\) Let's split this integral into two integrals: \(\int_{0}^{\pi} -\cos(x+a)\, dx + \int_{0}^{\pi} \cos{x}\, dx.\) For the first integral, we'll make a substitution. Let \(v = x+a\). Then, \(dv = dx\), and when \(x = 0\), \(v = a\), and when \(x = \pi\), \(v = \pi+a\). The integral becomes: \(-\int_{a}^{\pi+a} \cos(v)\, dv = -[\sin(v)]_{a}^{\pi+a} = -(\sin(\pi+a) - \sin{a})\). The second integral: \(\int_{0}^{\pi} \cos{x}\, dx = [\sin(x)]_{0}^{\pi} = (\sin(\pi) - \sin(0)) = 0\). Add both the integrals: \(-(\sin(\pi+a) - \sin{a}) + 0 = 1\).

Step 4: Solve for \(a\)

Now, we need to solve this equation to find the values of \(a\) in the given range \(0 \leq a \leq \pi\): \(\sin(\pi+a) - \sin{a} = 1\). Since \(\sin(\pi + a) = - \sin a\), we can rewrite the equation as: \(-2\sin{a} = 1\). Now, solving for \(a\) gives us: \(\sin{a} = -\frac{1}{2}\). In the range \(0 \leq a \leq \pi\), the value of \(a\) for which this equation holds true is: \(a = \frac{3\pi}{2}\).