Problem 30 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\sin x} \sin y d z d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral $\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\sin x} \sin y \, dz \, dx \, dy$. Answer: 4

Step 1: Integrate with respect to z

First, we need to integrate the function \(\sin y\) with respect to \(z\) within the bounds of \(0\) and \(\sin x\): $$ \int_{0}^{\sin x} \sin y \, dz $$ Since \(\sin y\) is constant with respect to \(z\), the integral is simply: $$ \sin y \int_{0}^{\sin x} dz = \sin y (z \Big|_{0}^{\sin x}) = \sin y (\sin x - 0) = \sin x \sin y $$ Now, we have a double integral: $$ \int_{0}^{\pi} \int_{0}^{\pi} \sin x \sin y \, dx \, dy $$

Step 2: Integrate with respect to x

Next, we need to integrate the function \(\sin x \sin y\) with respect to \(x\) within the bounds of \(0\) and \(\pi\): $$ \int_{0}^{\pi} \sin x \sin y \, dx $$ Since \(\sin y\) is constant with respect to \(x\), the integral becomes: $$ \sin y \int_{0}^{\pi} \sin x \, dx $$ We can integrate \(sin x\) with respect to \(x\): $$ \sin y \Big[-\cos x\Big|_0^{\pi} \Big] = \sin y (-\cos \pi + \cos 0) = 2 \sin y $$ Now, we have a single integral: $$ \int_{0}^{\pi} 2 \sin y \, dy $$

Step 3: Integrate with respect to y

Finally, we need to integrate the function \(2 \sin y\) with respect to \(y\) within the bounds of \(0\) and \(\pi\): $$ \int_{0}^{\pi} 2 \sin y \, dy $$ We can integrate \(2 \sin y\) with respect to \(y\): $$ \Big[-2 \cos y\Big|_0^{\pi} \Big] = -2 \cos \pi + 2 \cos 0 = 4 $$ So the final evaluation of the triple integral is: $$ \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\sin x} \sin y \, dz \, dx \, dy = 4 $$