/*! 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 5 Evaluate the iterated integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the iterated integral. $$ \int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{x} x y d y d x d z $$

Short Answer

Expert verified
The iterated integral evaluates to \(\frac{81}{5}\).

Step by step solution

01

Setup the Integral

We are given a triple integral with the limits of integration: the first integral (innermost) is with respect to \(y\), the second integral is with respect to \(x\), and the outermost integral is with respect to \(z\). Thus, we need to evaluate:\[ \int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{x} x y \, d y \, d x \, d z \]
02

Evaluate the Innermost Integral (with respect to y)

The innermost integral is with respect to \(y\) from 0 to \(x\). You integrate \(x y\) with respect to \(y\):\[ \int_{0}^{x} x y \, dy = x \cdot \frac{y^2}{2} \bigg|_0^x \]Substitute the limits:\[ x \cdot \frac{x^2}{2} - x \cdot \frac{0^2}{2} = \frac{x^3}{2} \]
03

Evaluate the Second Integral (with respect to x)

Next, integrate the result \(\frac{x^3}{2}\) with respect to \(x\) from 0 to \(\sqrt{9 - z^2}\):\[ \int_{0}^{\sqrt{9 - z^2}} \frac{x^3}{2} \, dx = \frac{1}{2} \cdot \frac{x^4}{4} \bigg|_0^{\sqrt{9 - z^2}} \]Substitute the limits:\[ \frac{1}{2} \cdot \frac{(9 - z^2)^2}{4} - 0 = \frac{(9 - z^2)^2}{8} \]
04

Evaluate the Outermost Integral (with respect to z)

Now, integrate \(\frac{(9 - z^2)^2}{8}\) with respect to \(z\) from 0 to 3:\[ \int_{0}^{3} \frac{(9 - z^2)^2}{8} \, dz \]The function \((9 - z^2)^2\) can be expanded to \(81 - 18z^2 + z^4\). The integral becomes:\[ \frac{1}{8} \left( \int_{0}^{3} 81 \, dz - 18 \int_{0}^{3} z^2 \, dz + \int_{0}^{3} z^4 \, dz \right)\]
05

Integrate each term

Now tackle each integral within the brackets:1. For \(\int_{0}^{3} 81 \, dz\): \[ 81z \bigg|_0^3 = 243 \]2. For \(18 \int_{0}^{3} z^2 \, dz\): \[ 18 \cdot \frac{z^3}{3} \bigg|_0^3 = 18 \cdot 9 = 162 \]3. For \(\int_{0}^{3} z^4 \, dz\): \[ \frac{z^5}{5} \bigg|_0^3 = \frac{243}{5} \]
06

Combine results and compute

Substitute these results back:\[ \frac{1}{8} \left(243 - 162 + \frac{243}{5} \right)\]Simplify inside the brackets:\[ 243 - 162 = 81 \]\[ 81 + \frac{243}{5} = \frac{405}{5} + \frac{243}{5} = \frac{648}{5}\]Finally, compute the result:\[ \frac{1}{8} \times \frac{648}{5} = \frac{648}{40} = \frac{81}{5}\]
07

Conclusion

The value of the iterated integral is \(\frac{81}{5}\).

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Key Concepts

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

Iterated Integrals
Iterated integrals are a systematic way of integrating a function of multiple variables over a specified region. In the context of triple integrals, the process involves evaluating multiple integrals successively.
This particular exercise illustrates the concept by presenting an iterated integral in the form \( \int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{x} x y \, d y \, d x \, d z \). Here, the integral is taken over three variables: \( y \), \( x \), and \( z \). Each layer of integration simplifies the function until all integrals are completed.
Understanding the order of integration is crucial as it dictates how each variable is handled. Always begin with the innermost integral, moving outward sequentially.
Iterated integrals are fundamental to solving problems in multivariable calculus, allowing us to compute volumes and other quantities in multi-dimensional spaces.
Integration Techniques
When dealing with iterated integrals, it is essential to apply the right integration techniques to efficiently evaluate each layer. This exercise provides a hands-on example:
  • Change of Variables: The bounds of the integrals and the replacement of variables based on limits represent a strategic plan that can simplify the problem.
  • Integration by Parts: Though not directly used in this simple problem, understanding this concept is key for more complex integrals.

In our solved exercise, the problem was approached by tackling the innermost integral. The function \( x y \) was integrated with respect to \( y \), yielding a result \( \frac{x^3}{2} \). Subsequently, further integrations were executed respecting the order of integration.
This structured approach ensures that each integration step simplifies the integral, making complex multivariable functions tractable.
Multivariable Calculus
Multivariable calculus extends the ideas of calculus to functions of several variables. This provides the tools necessary to analyze situations where numerous factors vary simultaneously.
In this exercise, we consider a function \( x y \) involving variables \( x \), \( y \), and \( z \). Calculating the triple integral of this function over a specific region in space provides valuable insights into volumetric properties and densities within that region.
  • Visualization: Imagine each allowed value of \( x \), \( y \), and \( z \) tracing a section within a larger 3D volume.
  • Applications: From finding the center of mass to electrical field calculations, multivariable calculus is used in various scientific and engineering disciplines.

By mastering triple integrals and iterated integrals in multivariable calculus, you gain the ability to solve complex, real-world problems involving multiple changing factors.

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Most popular questions from this chapter

Use a double integral in polar coordinates to find the area of the region described. The region enclosed by the cardioid \(r=1-\cos \theta\)

Find the mass and center of gravity of the lamina. A lamina with density \(\delta(x, y)=x^{2}+y^{2}\) is bounded by the \(x\) -axis and the upper half of the circle \(x^{2}+y^{2}=1\)

The tendency of a solid to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If the solid occupies a region \(G\) in an \(x y z\) -coordinate system, and if its density function \(\delta(x, y, z)\) is continuous on \(G,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iiint_{G}\left(y^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{y} &=\iiint_{G}\left(x^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{z} &=\iiint_{G}\left(x^{2}+y^{2}\right) \delta(x, y, z) d V \end{aligned}$$ In these exercises, find the indicated moments of inertia of the solid, assuming that it has constant density \(\delta .\) \(I_{z}\) for the solid cylinder \(x^{2}+y^{2} \leq a^{2}, 0 \leq z \leq h\)

The transformation \(x=a u, y=b v(a>0, b>0)\) can be rewritten as \(x / a=u, y / b=v,\) and hence it maps the circular region $$ u^{2}+v^{2} \leq 1 $$ into the elliptical region $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 $$ In these exercises, perform the integration by transforming the elliptical region of integration into a circular region of integration and then evaluating the transformed integral in polar coordinates. \(\iiint_{R} \sqrt{16 x^{2}+9 y^{2}} d A,\) where \(R\) is the region enclosed by the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 16\right)=1\)

Express the volume of the solid described as a double integral in polar coordinates. $$ \begin{array}{l}{\text { Inside of } x^{2}+y^{2}+z^{2}=9} \\ {\text { Outside of } x^{2}+y^{2}=1}\end{array} $$

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