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Chapter 16: Problem 8

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-1}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z d y d x d z$$

Short Answer

Expert verified
Based on the step-by-step solution provided above, create a short-answer question. Question: Evaluate the triple integral $$\int_{-1}^{1} \int_{-1}^{2} \int_{0}^{1} 6 x y z \, dy \, dx \, dz$$ and provide the resulting value. Answer: The value of the triple integral is 0.

Step by step solution

01

Evaluate the inner integral with respect to y

Integrate the function 6xyz with respect to y within the limits -1 to 2. $$\int_{-1}^{2} 6xyz\,dy = 6xz\int_{-1}^{2}y\,dy = 6xz\left[\frac{1}{2}y^2\right]_{-1}^{2} = 6xz(\frac{1}{2}(2^2) - \frac{1}{2}(-1)^2)$$
02

Simplify the result of Step 1

After simplifying the above result, we get: $$6xz(2 - \frac{1}{2}) = 6xz\cdot\frac{3}{2} = 9xz$$
03

Evaluate the middle integral with respect to x

Integrate the function 9xz with respect to x within the limits -1 to 1. $$\int_{-1}^{1} 9xz\,dx = 9z\int_{-1}^{1}x\,dx = 9z\left[\frac{1}{2}x^2\right]_{-1}^{1} = 9z(\frac{1}{2}(1^2) - \frac{1}{2}(-1)^2)$$
04

Simplify the result of Step 3

Since the second term and the first term of the result of step 3 are the same, they cancel each other out. Therefore, we have: $$9z(0) = 0$$
05

Evaluate the outer integral with respect to z

Integrate the function 0 with respect to z within the limits 0 to 1. $$\int_{0}^{1} 0\,dz = \int_{0}^{1} dz\,(0) = \left[0z\right]_{0}^{1}$$
06

Simplify the result of Step 5

The result is: $$0(1) - 0(0) = 0$$ The value of the triple integral is 0.

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Most popular questions from this chapter

Evaluate the following integrals using polar coordinates. Assume \((r, \theta)\) are polar coordinates. A sketch is helpful. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\}$$

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can, and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3} .\)

Improper integrals Many improper double integrals may be handled using the techniques for improper integrals in one variable (Section \(8.9) .\) For example, under suitable conditions on \(f\) $$ \int_{a}^{*} \int_{\varepsilon(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x $$ $$\int_{1}^{\infty} \int_{0}^{e^{-1}} x y d y d x$$

Suppose the density of a thin plate represented by the polar region \(R\) is \(\rho(r, \theta)\) (in units of mass per area). The mass of the plate is \(\iint_{R} \rho(r, \theta) d A .\) Find the mass of the thin half annulus \(R=\\{(r, \theta): 1 \leq r \leq 4,0 \leq \theta \leq \pi\\}\) with a density \(\rho(r, \theta)=4+r \sin \theta\).

Volume Find the volume of the solid bounded by the paraboloid \(z=2 x^{2}+2 y^{2},\) the plane \(z=0,\) and the cylinder \(x^{2}+(y-1)^{2}=1 .\) (Hint: Use symmetry.)
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