Chapter 16: Problem 22
Let \(W\) be the solid cone bounded by \(z=\sqrt{x^{2}+y^{2}}\) and \(z=2.\) Decide (without calculating its value) whether the integral is positive, negative, or zero. $$\int_{W} x d V$$
Short Answer
Expert verified
The integral is zero due to symmetry.
Step by step solution
01
Understand the Solid Region
The solid cone is bounded by the surface \(z = \sqrt{x^2 + y^2}\) and \(z = 2\). This is a cone with its vertex at the origin and opening upwards to the plane \(z = 2\).
02
Identify the Symmetry
The cone is symmetric around the \(z\)-axis. The function \(x\) is odd with respect to the \(y\)-axis, meaning that for every point \((x, y, z)\) in the region, there is a corresponding point \((-x, y, z)\) that will cancel it out due to symmetry.
03
Analyze the Integrand
The integrand \(x\) changes sign across the \(yz\)-plane (where \(x = 0\)). Due to the symmetry of the cone about the \(z\)-axis, every positive contribution of \(x\) will have a corresponding negative contribution, suggesting potential cancellation.
04
Consider the Resulting Integral
Given the symmetry of the region about the \(yz\)-plane and the nature of the function \(x\), the integral \(\int_{W} x \, dV\) sums up to zero as contributions to the integral cancel out evenly on either side of the \(yz\)-plane.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Triple Integrals
In multivariable calculus, triple integrals are a tool used to compute volumes and other properties over three-dimensional regions. When dealing with triple integrals, we're essentially integrating over a solid region in space. This means that instead of just thinking about a line or a plane, we consider a volume filled with points.
In the given example, we have a solid cone defined by boundaries on all three axes, hinting at the use of a triple integral. But, why use triple integrals? They allow us:
- To calculate the volume of a solid.
- To find total properties like mass, if the density is known across the region.
- To analyze functions over a three-dimensional space.
Volume of a Solid
The volume of a solid region, such as the cone described, can be determined using triple integrals. Since finding the exact volume of a more complex shape analytically can be tricky, using integration is a more rigorous method. In our case, the solid cone is easy to visualize: it grows upwards from its tip at the origin to a flat top at the plane given by the equation \(z = 2\). To find the volume of such a cone, you'd set up a triple integral with appropriate limits representing the dimensions of the region. You'd compute\[\int_{z = \sqrt{x^2+y^2}}^{z = 2} \, \int \int_{\text{base}} \, dV\]Where the base would be the circle obtained from \(z = \sqrt{x^2+y^2}\).However, in this problem, we are looking at whether a property of this solid is zero, not just its volume, due to the symmetry and properties of the integrand \(x\) over the region.
Symmetry in Integration
Symmetry plays a pivotal role in simplifying integrals, even though it might look intimidating initially. Symmetry tells us about the structure of a solid and how functions over it might behave.In the scenario of our cone, symmetry around the \(z\)-axis implies that every section of the solid on one side of the \(yz\)-plane (where \(x = 0\)) has a corresponding section on the other side. Because the integrand in our triple integral is \(x\), and \(x\) is an odd function with respect to the \(y\)-axis, it means:
- Positive values of \(x\) on one side of the \(yz\)-plane are mirrored by negative values.
- Each slice perpendicular to the \(x\)-axis contributes equally and oppositely to the total.
- The sum of these contributions across the entire volume results in cancellation.