/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Evaluate the following integrals... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
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 by step solution

01

First integral (x-integral)

Integrate \(\sin(\pi x)\) with respect to x over the interval [0, \(\pi/2\)]: $$\int_{0}^{\pi/2} \sin(\pi x) dx = \frac{1}{\pi} \int_{0}^{\pi/2} \sin(\pi x) (\pi dx) = \frac{1}{\pi} [-\cos(\pi x)]_{0}^{\pi/2} = \frac{1}{\pi}(-1 - 1) = -\frac{2}{\pi}$$
02

Second integral (y-integral)

Integrate \(\cos(y)\) with respect to y over the interval [0, 1]: $$\int_{0}^{1} \cos(y) dy = [\sin(y)]_{0}^{1} = \sin(1) - \sin(0) = \sin(1)$$
03

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: $$\int_{0}^{\pi / 2} \int_{0}^{1} \int_{0}^{\pi / 2} \sin(\pi x) \cos(y) \sin(2z) dy dx dz = -\frac{2\sin(1)}{\pi}$$

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