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Chapter 13: Problem 12

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

Short Answer

Expert verified
In this problem, the student is asked to evaluate the given triple integral, which represents the volume of a region. After following the steps to evaluate each integral, the final value of the triple integral is 3(e-1).

Step by step solution

01

Evaluate the integral with respect to x

Integrate the expression yze^x with respect to x over the interval [0, 1]: $$\int_{0}^{1} y z e^{x} dx.$$ Using the integration rule for exponential functions, we get: $$[yze^x]_{0}^{1} = yze^1 - yze^0$$ The result is: $$yz(e-1)$$
02

Evaluate the integral with respect to z

Now, integrate the expression yz(e-1) with respect to z over the interval [1, 2]: $$\int_{1}^{2} yz(e-1) dz$$ Using the power rule for integration, we get: $$\left[\frac{1}{2}yz^2(e-1)\right]_{1}^{2} = y\left(\frac{1}{2}(2^2 - 1^2)(e-1)\right) = \frac{3}{2}y(e-1)$$
03

Evaluate the integral with respect to y

Finally, integrate the expression \(\frac{3}{2}y(e-1)\) with respect to y over the interval [0, 2]: $$\int_{0}^{2} \frac{3}{2}y(e-1) dy$$ Using the power rule again, we get: $$\left[\frac{1}{2}\cdot\frac{3}{2}y^2(e-1)\right]_{0}^{2} = \frac{3}{4}(4)(e-1) = 3(e-1)$$ The value of the triple integral is: $$\int_{0}^{2} \int_{1}^{2} \int_{0}^{1} y z e^{x} d x d z d y = 3(e-1)$$

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Most popular questions from this chapter

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\begin{array}{l} \iint_{R} \frac{d A}{4+\sqrt{x^{2}+y^{2}}} ; R=\\{(r, \theta): 0 \leq r \leq 2 \\\ \pi / 2 \leq \theta \leq 3 \pi / 2\\} \end{array}$$

Spherical coordinates Evaluate the Jacobian for the transformation from spherical to rectangular coordinates: \(x=\rho \sin \varphi \cos \theta, y=\rho \sin \varphi \sin \theta, z=\rho \cos \varphi .\) Show that \(J(\rho, \varphi, \theta)=\rho^{2} \sin \varphi\)

Miscellaneous volumes Choose the best coordinate system for finding the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. That part of the solid cylinder \(r \leq 2\) that lies between the cones \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\)

Miscellaneous volumes Choose the best coordinate system for finding the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. That part of the ball \(\rho \leq 2\) that lies between the cones \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\)

Let \(f\) be a continuous function on \([0,1] .\) Prove that $$\int_{0}^{1} \int_{x}^{1} \int_{x}^{y} f(x) f(y) f(z) d z d y d x=\frac{1}{6}\left(\int_{0}^{1} f(x) d x\right)^{3}$$
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