Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 16
\(\iiint_{E} z d V,\) where \(E\) is bounded by the cylinder \(y^{2}+z^{2}=9\) and the planes \(x=0, y=3 x,\) and \(z=0\) in the first octant
Short Answer
Expert verified
The value of the triple integral is 0.
Step by step solution
01
Analyze the region E
The region \( E \) is bounded by several constraints. The cylinder \( y^2 + z^2 = 9 \) describes a vertical cylinder centered along the x-axis with a radius of 3. The plane \( x = 0 \) means the region starts from the yz-plane. The plane \( y = 3x \) limits the region in the xy-plane, indicating the boundary is a plane passing through the origin and with a slope of 3 in the xy-plane. Finally, \( z = 0 \) is the xy-plane itself, bounding \( z \) from below, while z is bounded above by the cylinder.
02
Set up the bounds for integration
Since the region is in the first octant, all coordinates are non-negative. The limits for \( y \) and \( z \) come from the cylinder equation \( y^2 + z^2 = 9 \). In terms of \( x \), as \( y = 3x \), \( x \) will vary from 0 to \( x = y/3 \). Thus, the bounds for \( x \) are \( 0 \leq x \leq y/3 \), for \( y \) from 0 to 3, and for \( z \) from 0 to \( \sqrt{9 - y^2} \), since \( z \) is limited by the cylinder.
03
Write the triple integral
The integral can now be expressed in the order \( dz \), \( dy \), \( dx \) as: \[ \int_{0}^{3} \int_{0}^{\sqrt{9-y^2}} \int_{0}^{y/3} z \, dx \, dz \, dy \]
04
Integrate with respect to x
Integrate the innermost integral: \[ \int_{0}^{y/3} z \, dx = zx \Big|_{0}^{y/3} = z \left(\frac{y}{3}\right) = \frac{yz}{3} \] Thus, the updated integral is: \[ \int_{0}^{3} \int_{0}^{\sqrt{9-y^2}} \frac{yz}{3} \, dz \, dy \]
05
Integrate with respect to z
Integrate the result with respect to \( z \): \[ \int_{0}^{\sqrt{9-y^2}} \frac{yz}{3} \, dz = \frac{y}{3} \left[ \frac{z^2}{2} \right]_{0}^{\sqrt{9-y^2}} = \frac{y}{3} \cdot \frac{9-y^2}{2} \] This simplifies to: \[ \frac{y(9-y^2)}{6} \] The integral is now: \[ \int_{0}^{3} \frac{y(9-y^2)}{6} \, dy \]
06
Integrate with respect to y
Expand and integrate with respect to \( y \): \[ \frac{1}{6} \int_{0}^{3} (9y - y^3) \, dy = \frac{1}{6} \left[ \frac{9y^2}{2} - \frac{y^4}{4} \right]_{0}^{3} \]Calculate: \[ = \frac{1}{6} \left( \frac{9 \times 9}{2} - \frac{81}{4} \right) \]\[ = \frac{1}{6} \left( \frac{81 - 81}{4} \right) = 0 \]
07
Conclusion
The resulting value of the triple integral, which represents the volume-weighted sum, is 0. This occurs as the integration suggests the region defined contributes an overall zero weight when considering the plane \( z = 0 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
cylindrical coordinates
Cylindrical coordinates are a system used in mathematics to describe 3-dimensional space. This system is especially handy when dealing with problems that involve symmetry around an axis, like cylinders or anything that wraps around a central point.
Cylindrical coordinates consist of three values:
Cylindrical coordinates consist of three values:
- The radial distance \( r \), which is the perpendicular distance from the point to the z-axis.
- The angle \( \theta \), which is the counterclockwise angle in the xy-plane from the positive x-axis.
- The height \( z \), which is the same as in Cartesian coordinates, indicating the height above or below the xy-plane.
volumes of regions
Understanding the volumes of regions is crucial in problems involving triple integrals. In this context, a triple integral calculates the volume of a three-dimensional region, which may be bounded or confined by various surfaces or planes.
The original exercise involves calculating the volume of the region bounded by a cylinder and several planes in the first octant. The first octant constraint ensures that all coordinates \( x, y, \) and \( z \) are non-negative. You can imagine this as a pie-slice within the 3D space.
Setting up the bounds of our integrals is the first crucial step. You determine:
The original exercise involves calculating the volume of the region bounded by a cylinder and several planes in the first octant. The first octant constraint ensures that all coordinates \( x, y, \) and \( z \) are non-negative. You can imagine this as a pie-slice within the 3D space.
Setting up the bounds of our integrals is the first crucial step. You determine:
- Where the region starts and ends in terms of \( x \), \( y \), and \( z \).
- As the question highlights, the region of integration is defined by the cylinder, which can be thought of as limiting \( y \) and \( z \).
- The planes provide boundaries for the \( x \) coordinate in the context of this bounded region.
coordinate planes
Coordinate planes are fundamental in understanding and solving problems in 3D space. They refer to the xyz-coordinate system planes: the xy-plane, yz-plane, and xz-plane, acting as reference sheets slicing through the 3D space.
In the given exercise, the planes mentioned are \( x = 0 \), \( y = 3x \), and \( z = 0 \):
In the given exercise, the planes mentioned are \( x = 0 \), \( y = 3x \), and \( z = 0 \):
- \( x = 0 \): The yz-plane represents the starting boundary for the region we are investigating. It is like setting the leftmost part of a 3D coordinate system.
- \( y = 3x \): This plane limits the region in the xy-plane. Its slope tells you how \( y \) grows in relation to \( x \). Therefore, as you travel along \( x \), \( y \) expands approximately three times faster until hitting the boundary.
- \( z = 0 \): The xy-plane restricts the region's lower bound in the z-direction, making the whole region fall above the xy-plane.
