Chapter 15: Problem 18
Evaluate the triple integral. $$ \begin{array}{l}{\iint_{E} z d V, \text { where } E \text { is bounded by the cylinder } y^{2}+z^{2}=9} \\ {\text { and the planes } x=0, y=3 x, \text { and } z=0 \text { in the first octant }}\end{array} $$
Short Answer
Expert verified
The value of the triple integral is \( \frac{27}{8} \).
Step by step solution
01
Understand the Region of Integration
The region of integration, \(E\), is bounded by the cylinder \(y^2 + z^2 = 9\) and the planes \(x=0\), \(y = 3x\), and \(z = 0\) in the first octant. The region is in the positive section of the 3-dimensional space limited by the given cylinder and planes.
02
Determine Limits for x-coordinate
The plane \(y = 3x\) provides a relationship between \(y\) and \(x\). Since \(x\) is in the first octant, it begins at 0 and ends where it intersects the cylinder. In terms of \(x\), this corresponds to points where \(y = 3x = 3\sqrt{9 - z^2}\), thus giving \(x\) limits: \(0 \leq x \leq \frac{y}{3}\).
03
Determine Limits for y-coordinate
Since \(y^2 + z^2 = 9\), the maximum value for \(y\) in the positive axis must be \(3\) (the circle's radius). Therefore, \(0 \leq y \leq 3\), limited by the intersection of \(y = 3x\).
04
Determine Limits for z-coordinate
The cylinder \(y^2 + z^2 = 9\) defines a circular base. The \(z\)-coordinate is on this circle from 0 to \(\sqrt{9 - y^2}\). Thus, \(0 \leq z \leq \sqrt{9 - y^2}\).
05
Write the Triple Integral
Given the constraints, the triple integral is set up as: \[\int_{0}^{3} \int_{0}^{\frac{y}{3}} \int_{0}^{\sqrt{9 - y^2}} z \, dz \, dx \, dy. \]
06
Evaluate the Innermost Integral
First, integrate with respect to \(z\): \[ \int_{0}^{\sqrt{9 - y^2}} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{\sqrt{9 - y^2}} = \frac{(9 - y^2)}{2}. \]
07
Evaluate the Middle Integral
Next, integrate with respect to \(x\): \[\int_{0}^{y/3} \frac{(9-y^2)}{2} \, dx = \frac{(9-y^2)}{2} \left[ x \right]_{0}^{y/3} = \frac{(9-y^2)x}{2} \Bigg|_{0}^{y/3} = \frac{(9-y^2)y}{6}. \]
08
Evaluate the Outer Integral
Finally, integrate with respect to \(y\): \[\int_{0}^{3} \frac{(9-y^2)y}{6} \, dy = \frac{1}{6} \left[ \frac{9y^2}{2} - \frac{y^4}{4} \right]_{0}^{3} = \frac{1}{6} \left( \frac{9 \cdot 9}{2} - \frac{81}{4} \right).\]After simplifying, you'll find the result to be \( \frac{27}{8} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylindrical Coordinates
Cylindrical coordinates are a powerful alternative to Cartesian coordinates when dealing with problems that have symmetry about an axis, such as when dealing with cylinders. In cylindrical coordinates, we use a combination of radius, angle, and height to locate points in space. This system is especially useful in three-dimensional space when dealing with circular or cylindrical regions.
The core elements of cylindrical coordinates are:
The core elements of cylindrical coordinates are:
- Radial distance ( ) from the origin to the projection of the point in the xy-plane.
- Angular coordinate ( heta) as the angle from the positive x-axis to the projection of the point in the xy-plane.
- Height (z) above the xy-plane.
Volume Integration
Triple integrals are used to calculate volumes, among other applications. They extend the concept of double integrals to three dimensions. Volume integration over a region allows you to compute accumulated quantities, such as mass or charge, throughout a 3D region.
To evaluate a triple integral, \( \int \int \int_{E} f(x, y, z) \, dV \), we:
To evaluate a triple integral, \( \int \int \int_{E} f(x, y, z) \, dV \), we:
- Establish functions or surfaces that bound the solid, defining a region \(E\).
- Determine the limits for each of the three coordinates, usually set from the simplest to most complex integral.
- Integrate the function, starting from the innermost integral moving outwards.
First Octant
The first octant in three-dimensional space refers to the portion where all the Cartesian coordinates \((x, y, z)\) are positive. This quadrant is vital in physics and engineering, representing a complete set of possibilities for positive dimension values. Declaring the first octant simplifies the analysis because:
- It assures positive values, eliminating the need to consider negative plane intersections.
- It sets clear boundaries, aiding in visualizing and calculating integrals for problems within this space.