Problem 11 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{0}^{\pi / 2} \int_{0}^{1} \int_{0}^{\pi / 2} \sin \pi x \cos y \sin 2 z d y d x d z$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral $$\int_{0}^{\pi / 2} \int_{0}^{1} \int_{0}^{\pi / 2} \sin(\pi x) \cos(y) \sin(2z) dy dx dz$$ Answer: $$-\frac{2\sin(1)}{\pi}$$

Step 1: First integral (x-integral)

Integrate \(\sin(\pi x)\) with respect to x over the interval [0, \(\pi/2\)]: $$\int_{0}^{\pi/2} \sin(\pi x) dx = \frac{1}{\pi} \int_{0}^{\pi/2} \sin(\pi x) (\pi dx) = \frac{1}{\pi} [-\cos(\pi x)]_{0}^{\pi/2} = \frac{1}{\pi}(-1 - 1) = -\frac{2}{\pi}$$

Step 2: Second integral (y-integral)

Integrate \(\cos(y)\) with respect to y over the interval [0, 1]: $$\int_{0}^{1} \cos(y) dy = [\sin(y)]_{0}^{1} = \sin(1) - \sin(0) = \sin(1)$$

Step 3: Third integral (z-integral)

Integrate \(\sin(2z)\) with respect to z over the interval [0, \(\pi/2\)]: $$\int_{0}^{\pi/2} \sin(2z) dz = \frac{1}{2} \int_{0}^{\pi/2} \sin(2z) (2 dz) = \frac{1}{2} [-\cos(2z)]_{0}^{\pi/2} = \frac{1}{2} (1 - (-1)) = 1$$ Now, we multiply all three integral results together: $$-\frac{2}{\pi} \cdot \sin(1) \cdot 1 = -\frac{2\sin(1)}{\pi}$$ Thus, the value of the triple integral is: $$\int_{0}^{\pi / 2} \int_{0}^{1} \int_{0}^{\pi / 2} \sin(\pi x) \cos(y) \sin(2z) dy dx dz = -\frac{2\sin(1)}{\pi}$$