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Chapter 16: Problem 11

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
In this problem, we evaluated a triple integral with the given limits of integration and the integrand function \(\sin (\pi x) \cos y \sin 2z\). By integrating with respect to \(y\) first, then \(x\), and finally \(z\), we found that the value of the triple integral is \(\frac{1}{\pi}\).

Step by step solution

01

Integrate with respect to y

First, we will integrate the integrand function with respect to \(y\): $$\int_0^{\pi / 2} \sin \pi x \cos y \sin 2 z dy$$ Since \(\sin \pi x\) and \(\sin 2z\) are constants with respect to \(y\), we can move them outside the integral: $$\sin \pi x \sin 2z \int_0^{\pi / 2} \cos y dy$$ Now, we can integrate \(\cos y\) with respect to \(y\): $$\sin \pi x \sin 2z [\sin y]_0^{\pi / 2}$$ $$\sin \pi x \sin 2z(\sin (\pi / 2) - \sin 0)$$ This simplifies to: $$\sin \pi x \sin 2z$$
02

Integrate with respect to x

Now, we will integrate the result from Step 1 with respect to \(x\): $$\int_0^1 \sin \pi x \sin 2z dx$$ Since \(\sin 2z\) is a constant with respect to \(x\), we can move it outside the integral: $$\sin 2z \int_0^1 \sin \pi x dx$$ Now, we can integrate \(\sin \pi x\) with respect to \(x\): $$\sin 2z \left[-\frac{1}{\pi}\cos(\pi x)\right]_0^1$$ $$\sin 2z\left(-\frac{1}{\pi}(\cos(\pi) - \cos(0))\right)$$ This simplifies to: $$\frac{2}{\pi}\sin 2z$$
03

Integrate with respect to z

Finally, we will integrate the result from Step 2 with respect to \(z\): $$\int_0^{\pi / 2} \frac{2}{\pi} \sin 2z dz$$ Now, we can integrate \(\sin 2z\) with respect to \(z\): $$\frac{2}{\pi}\left[-\frac{1}{2} \cos(2z)\right]_0^{\pi / 2}$$ $$-\frac{1}{\pi}\left(\cos(\pi) - \cos(0)\right)$$ This simplifies to: $$\frac{1}{\pi}$$ So the value of the triple integral is: $$\int_{0}^{\pi / 2} \int_{0}^{1} \int_{0}^{\pi / 2} \sin \pi x \cos y \sin 2z dy dx dz = \frac{1}{\pi}$$

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