/*! 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 1 \(\int_{2}^{4} \int_{-2}^{2} \in... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 1

\(\int_{2}^{4} \int_{-2}^{2} \int_{-1}^{1}(x+y+z) d x d y d z=\left.\int_{2}^{4} \int_{-2}^{2}\left(\frac{1}{2} x^{2}+x y+x z\right)\right|_{-1} ^{1} d y d z\) $$=\int_{2}^{4} \int_{-2}^{2}(2 y+2 z) d y d z=\left.\int_{2}^{4}\left(y^{2}+2 y z\right)\right|_{-2} ^{2} d z=\int_{2}^{4} 8 z d z=\left.4 z^{2}\right|_{2} ^{4}=48$$

Short Answer

Expert verified
The triple integral evaluates to 48.

Step by step solution

01

Understand the Problem

We are given a triple integral \[\int_{2}^{4} \int_{-2}^{2} \int_{-1}^{1}(x+y+z) \ d x \ d y \ d z\] that needs to be evaluated.
02

Integrate with respect to x

First, integrate the function \(x+y+z\) with respect to \(x\) from \(-1\) to \(1\) while keeping \(y\) and \(z\) as constants:\[\int_{-1}^{1} (x+y+z) \, dx = \left[ \frac{1}{2} x^2 + xy + xz \right]_{-1}^{1}\]
03

Evaluate x-integral limits

Substitute the limits for \(x\) into the integrated function:Evaluate at \(x = 1\):\[\frac{1}{2}(1)^2 + (1)y + (1)z = \frac{1}{2} + y + z\]Evaluate at \(x = -1\):\[\frac{1}{2}(-1)^2 + (-1)y + (-1)z = \frac{1}{2} - y - z\]Subtract these results:\[\left(\frac{1}{2} + y + z\right) - \left(\frac{1}{2} - y - z\right) = 2y + 2z\]
04

Integrate with respect to y

Now integrate \(2y + 2z\) with respect to \(y\) from \(-2\) to \(2\):\[\int_{-2}^{2} (2y + 2z) \, dy = \left[ y^2 + 2yz \right]_{-2}^{2}\]
05

Evaluate y-integral limits

Substitute the limits for \(y\):Evaluate at \(y = 2\):\[(2)^2 + 2(2)z = 4 + 4z\]Evaluate at \(y = -2\):\[(-2)^2 + 2(-2)z = 4 - 4z\]Subtract these results:\[4 + 4z - (4 - 4z) = 8z\]
06

Integrate with respect to z

Now integrate \(8z\) with respect to \(z\) from \(2\) to \(4\):\[\int_{2}^{4} 8z \, dz = \left[ 4z^2 \right]_{2}^{4}\]
07

Evaluate z-integral limits

Substitute the limits for \(z\):Evaluate at \(z = 4\):\[4(4)^2 = 64\]Evaluate at \(z = 2\):\[4(2)^2 = 16\]Subtract these results:\[64 - 16 = 48\]
08

Final Result

After evaluating all the integrals and subtracting the limits, the value of the original triple integral is 48.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
A definite integral is a key concept in calculus used to calculate the area under a curve between two specific points on the real line. For a function \( f(x) \), the definite integral from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \).
This integral provides a number that represents the cumulative effect of the function over the interval \([a, b]\).
  • To compute a definite integral, one finds an antiderivative (or indefinite integral) of the function.
  • Next, evaluate this antiderivative at the upper and lower limits of the interval \(a\) and \(b\).
  • The final step is to subtract these two evaluations.
For example, in the exercise, we are dealing with a triple integral, which essentially combines three definite integrals over the three variables \(x, y,\) and \(z\). Each integral corresponds to one dimension and is evaluated separately starting from the innermost (with respect to \(x\)) to the outermost (with respect to \(z\)) limits.
Iterated Integrals
Iterated integrals refer to the repetitive integration process performed when dealing with multiple variables, such as in double or triple integrals. In this structured approach, integrals are calculated one after the other.
  • Start with the innermost integral: When dealing with triple integrals, the function inside is first integrated with respect to the innermost variable, holding other variables constant.
  • Proceed to the next integral: Once the first integration is completed, calculate the next integral with respect to the second variable.
  • Finish with the outermost integral: The process ends with the evaluation of the last integral with respect to the outermost variable.
In the given exercise, the integration was performed iteratively:- Integration began with \( \int_{-1}^{1} (x+y+z) \, dx \), holding \(y\) and \(z\) constant.- The result \(2y + 2z\) was obtained, and the next integration \( \int_{-2}^{2} (2y + 2z) \, dy \) was performed.- Finally, the outermost integration \( \int_{2}^{4} 8z \, dz \) was evaluated to arrive at the result.
Calculus Tutorial
Calculus, the branch of mathematics that studies continuous change, offers powerful tools such as derivatives and integrals. For students delving into triple integrals, it鈥檚 helpful to consider the foundational aspects first.
  • Understand the basics: A good grasp of single-variable calculus is essential before tackling multivariable calculus. Definite and indefinite integrals should be mastered first.
  • Familiarize with multivariable functions: Learn how to work with functions of more than one variable and understand how to hold some variables constant while manipulating others.
  • Practice iterated integrals: Start by solving problems with double integrals before moving on to the complexity of triple integrals.
By methodically approaching the subject, students can gain confidence. - By understanding the step-by-step solutions, students learn how to systematically approach each layer of integration. - This practice provides insight into how multivariable functions integrate in a defined region of space.

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Most popular questions from this chapter

The other five integrals are \(\int_{0}^{3} \int_{0}^{\sqrt{36-4 y^{2}} / 3} \int_{1}^{3} F(x, y, z) d z d x d y\), $$\begin{aligned} &\int_{1}^{3} \int_{0}^{2} \int_{0}^{\sqrt{36-9 x^{2} / 2}} F(x, y, z) d y d x d z, \int_{1}^{3} \int_{0}^{3} \int_{0}^{\sqrt{36-4 y^{2}} / 3} F(x, y, z) d x d y d z,\\\ &\int_{0}^{3} \int_{1}^{3} \int_{0}^{\sqrt{36-4 y^{2}}} F(x, y, z) d x d z d y, \int_{0}^{2} \int_{1}^{3} \int_{0}^{\sqrt{36-9 x^{2}} / 2} F(x, y, z) d y d z d x. \end{aligned}$$

The equation of the ellipse is \(9 x^{2} / a^{2}+4 y^{2} / b^{2}=1\) and the equation of the parabola is \(y=\pm\left(9 b x^{2} / 8 a^{2}-b / 2\right)\). Letting \(I_{e}\) and \(I_{p}\) represent the moments of inertia of the ellipse and parabola, respectively, about the \(x\) -axis, we have \(I_{e}=2 \int_{-a / 3}^{0} \int_{0}^{b \sqrt{a^{2}-9 x^{2}} / 2 a} y^{2} d y d x=\frac{b^{3}}{12 a^{3}} \int_{-a / 3}^{0}\left(a^{2}-9 x^{2}\right)^{3 / 2} d x\) \(x=\frac{a}{3} \sin \theta, \quad d x=\frac{a}{3} \cos \theta d \theta\) \(=\frac{b^{3}}{12 a^{3}} \frac{a^{4}}{3} \int_{-\pi / 3}^{0} \cos ^{4} \theta d \theta=\frac{b^{3} a}{36} \frac{3 \pi}{16}=\frac{a b^{3} \pi}{192}\) and \(I_{p}=2 \int_{0}^{2 a / 3} \int_{0}^{b / 2-9 b x^{2} / 8 a^{2}} y^{2} d y d x=\frac{2}{3} \int_{0}^{2 a / 3}\left(\frac{b}{2}-\frac{9 b}{8 a^{2}} x^{2}\right)^{3} d x=\frac{2}{3} \frac{b^{3}}{8} \int_{0}^{2 a / 3}\left(1-\frac{9}{4 a^{2}} x^{2}\right)^{3} d x\) \(=\frac{b^{3}}{12} \int_{0}^{2 a / 3}\left(1-\frac{27}{4 a^{2}} x^{2}+\frac{243}{16 a^{4}} x^{4}-\frac{729}{64 a^{6}} x^{6}\right) d x=\left.\frac{b^{3}}{12}\left(x-\frac{9}{4 a^{2}} x^{3}+\frac{243}{80 a^{4}} x^{5}-\frac{729}{64 a^{6}} x^{7}\right)\right|_{0} ^{2 a / 3}\) \(=\frac{b^{3}}{12} \frac{32 a}{105}=\frac{8 a b^{3}}{315}\). Then \(I_{x}=I_{e}+I_{p}=\frac{a b^{3} \pi}{192}+\frac{8 a b^{3}}{315}\).

\(R 1: y=-3 x+3 \Longrightarrow v=3\) \(R 2: y=x-\pi \Longrightarrow u=\pi\) \(R 3: y=-3 x+6 \Longrightarrow v=6\) \(R 4: y=x \Longrightarrow u=0\) \(\frac{\partial(u, v)}{\partial(x, y)}=\left|\begin{array}{cc}1 & -1 \\ 3 & 1\end{array}\right|=4 \Rightarrow \frac{\partial(x, y)}{\partial(u, v)}=\frac{1}{4}\) $$\begin{aligned} \iint_{R} \frac{\cos \frac{1}{2}(x-y)}{3 x+y} d A &=\iint_{S} \frac{\cos u / 2}{v}\left(\frac{1}{4}\right) d A^{\prime}=\frac{1}{4} \int_{3}^{6} \int_{0}^{\pi} \frac{\cos u / 2}{v} d u d v=\left.\frac{1}{4} \int_{3}^{6} \frac{2 \sin u / 2}{v}\right|_{0} ^{\pi} d v \\ &=\frac{1}{2} \int_{3}^{6} \frac{d v}{v}=\left.\frac{1}{2} \ln v\right|_{3} ^{6}=\frac{1}{2} \ln 2 \end{aligned}$$

$$\text { With } x=-5, y=-5, \text { and } z=0, \text { we have } \rho^{2}=50, \tan \theta=1, \text { and } \cos \phi=0 . \text { The point is }(5 \sqrt{2}, \pi / 2,5 \pi / 4)$$.

\(m=\int_{0}^{1} \int_{0}^{e^{x}} y^{3} d y d x=\left.\int_{0}^{1} \frac{1}{4} y^{4}\right|_{0} ^{e^{x}} d x=\int_{0}^{1} \frac{1}{4} e^{4 x} d x=\left.\frac{1}{16} e^{4 x}\right|_{0} ^{1}=\frac{1}{16}\left(e^{4}-1\right)\) \(M_{y}=\int_{0}^{1} \int_{0}^{e^{x}} x y^{3} d y d x=\left.\int_{0}^{1} \frac{1}{4} x y^{4}\right|_{0} ^{e^{x}} d x=\int_{0}^{1} \frac{1}{4} x e^{4 x} d x\) Integration by parts \(=\left.\frac{1}{4}\left(\frac{1}{4} x e^{4 x}-\frac{1}{16} e^{4 x}\right)\right|_{0} ^{1}=\frac{1}{4}\left(\frac{3}{16} e^{4}+\frac{1}{16}\right)=\frac{1}{64}\left(3 e^{4}+1\right)\) \(M_{x}=\int_{0}^{1} \int_{0}^{e^{x}} y^{4} d y d x=\left.\int_{0}^{1} \frac{1}{5} y^{5}\right|_{0} ^{e^{x}} d x=\int_{0}^{1} \frac{1}{5} e^{5 x} d x=\left.\frac{1}{25} e^{5 x}\right|_{0} ^{1}=\frac{1}{25}\left(e^{5}-1\right)\) \(\bar{x}=M_{y} / m=\frac{\left(3 e^{4}+1\right) / 64}{\left(e^{4}-1\right) / 16}=\frac{3 e^{4}+1}{4\left(e^{4}-1\right)} ; \quad \bar{y}=M_{x} / m=\frac{\left(e^{5}-1\right) / 25}{\left(e^{4}-1\right) / 16}=\frac{16\left(e^{5}-1\right)}{25\left(e^{4}-1\right)}\) The center of mass is \(\left(\frac{3 e^{4}+1}{4\left(e^{4}-1\right)}, \frac{16\left(e^{5}-1\right)}{25\left(e^{4}-1\right)}\right) \approx(0.77,1.76)\).
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