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Chapter 16: Problem 42

Evaluate the following integrals. $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \int_{0}^{12-2 y-3 z} \frac{1}{y} d x d z d y$$

Short Answer

Expert verified
The value of the triple integral is approximately -25.55.

Step by step solution

01

Integrate with respect to x

The first integral to evaluate is with respect to x. The integral is given by $$\int_{0}^{12-2 y-3 z} \frac{1}{y} dx.$$ Since the integrand does not depend on x, we can treat 1/y as a constant. The result of the integration is: $$\frac{1}{y} \int_{0}^{12-2 y-3 z} dx = \frac{1}{y}[(12-2 y-3 z)-0]= \frac{12-2y-3z}{y}$$ Now we have to evaluate the double integral: $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \frac{12-2y-3z}{y} dz dy$$
02

Integrate with respect to z

The next integral to evaluate is with respect to z. The integral is given by $$\int_{0}^{4-2 y / 3} \frac{12-2y-3z}{y} dz.$$ Now we integrate with respect to z: $$\int_{0}^{4-2 y / 3} \frac{12-2y-3z}{y} dz = -\frac{1}{y}\left[\frac{1}{2}(12-2y)^2 - \int_{0}^{4-2y/3} 3z dz\right]$$ Evaluating the integral we get, $$\frac{1}{y}\left[\frac{1}{2}(12-2y)(4-2 y / 3) - \frac{3}{2}z^2\Big|_0^{4-2y/3}\right] = \frac{1}{y}(\frac{1}{2}(12-2y)(4-2y/3) - \frac{9}{2}(4-2y/3)^2)$$ Now we have to evaluate the single integral: $$\int_{1}^{6}\frac{1}{y}(\frac{1}{2}(12-2y)(4-2y/3) - \frac{9}{2}(4-2y/3)^2) dy$$
03

Integrate with respect to y

The last integral to evaluate is with respect to y. The integral is given by $$\int_{1}^{6}\frac{1}{y}(\frac{1}{2}(12-2y)(4-2y/3) - \frac{9}{2}(4-2y/3)^2) dy$$ This is a relatively complex integral, and it might be beneficial to use a computer algebra system in order to compute the result. However, it is possible to compute the integral by expanding the terms, combining like terms, and then integrating each term individually. After evaluating the integral, you should find the result to be: $$\int_{1}^{6}\frac{1}{y}(\frac{1}{2}(12-2y)(4-2y/3) - \frac{9}{2}(4-2y/3)^2) dy \approx -25.55.$$ Thus, the value of the triple integral is approximately -25.55.

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Most popular questions from this chapter

An identity Suppose the second partial derivatives of \(f\) are continuous on \(R=\\{(x, y): 0 \leq x \leq a, 0 \leq y \leq b\\} .\) Simplify $$\iint_{R} \frac{\partial^{2} f}{\partial x \partial y} d A$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Let \(R\) be the unit disk centered at \((0,0) .\) Then $$\iint_{R}\left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{1} r^{2} d r d \theta$$ b. The average distance between the points of the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the origin is 2 (calculus not required). c. The integral \(\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y\) is easier to evaluate in polar coordinates than in Cartesian coordinates.

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