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Chapter 5: Problem 197

In the following exercises, evaluate the triple integrals over the indicated bounded region \(E\). \(\quad\) ? \(y d V, \quad\) where \(E=\left\\{(x, y, z) \mid-1 \leq x \leq 1,-\sqrt{1-x^{2}} \leq y \leq \sqrt{1-x^{2}}, 0 \leq z \leq 1-x^{2}-y^{2}\right\\}\)

Short Answer

Expert verified
The value of the triple integral is 0.

Step by step solution

01

Understand the Region

The region \(E\) is defined in a cylindrical form with \(x\) ranging from \(-1\) to \(1\). For each \(x\), \(y\) varies from \(-\sqrt{1-x^2}\) to \(\sqrt{1-x^2}\), which describes a circular cross-section in the \(xy\)-plane with unit radius. The height \(z\) goes from \(0\) to \(1-x^2-y^2\), forming a paraboloid that is bounded above by the unit sphere.
02

Set Up the Triple Integral

The integral in terms of \(dV = dy\,dx\,dz\) can be expressed as: \[ \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{1-x^2-y^2} y \ dz \ dy \ dx \] Here, the innermost integral is with respect to \(z\), the middle integral is with respect to \(y\), and the outermost integral is with respect to \(x\).
03

Integrate with Respect to z

The first integration is with respect to \(z\): \[ \int_{0}^{1-x^2-y^2} y \, dz = y \cdot z \bigg|_{0}^{1-x^2-y^2} = y(1-x^2-y^2) \] Thus, the integrated expression becomes \( y(1-x^2-y^2) \).
04

Integrate with Respect to y

Now, integrate the expression \( y(1-x^2-y^2) \) with respect to \(y\):\[ \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y(1-x^2-y^2) \ dy \] Split and solve:\[ = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y - x^2y - y^3 \, dy \]Evaluate the terms:\[ = \left[ \frac{y^2}{2} - \frac{x^2y^2}{2} - \frac{y^4}{4} \right]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \]Since \(y\) is symmetric about zero and \(x^2y^2\) and \(y^4\) are even functions, each integral over symmetric limits evaluates to zero, simplifying our calculation.
05

Evaluate the y Integral

Substitute the limits \(y = \sqrt{1-x^2}\) and \(y = -\sqrt{1-x^2}\):\[ \frac{(1-x^2)}{2} \cdot 1 \] This simplifies to zero because \(- (\sqrt{1-x^2})^4 \) and \((\sqrt{1-x^2})^4 \) cancel each other, resulting in 0.
06

Integrate with Respect to x

The integral with respect to \(x\) becomes:\[ \int_{-1}^{1} 0 \ dx = 0 \] Since the expression is zero, the integration over \(x\) results in zero.

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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 natural extension of polar coordinates into three dimensions, particularly useful when dealing with symmetrical objects or regions like cylinders. This system uses:
  • The radial distance \( r \), which is the distance from the origin to the projection of a point in the \( xy \)-plane.
  • The angle \( \theta \), which is the angle formed with the positive \( x \)-axis.
  • The height \( z \), representing vertical positioning parallel to the \( z \)-axis.
For any point \((x, y, z)\), the transformation equations to convert from cylindrical coordinates to Cartesian coordinates are:
\[ x = r \cos \theta \] \[ y = r \sin \theta \] \[ z = z \] This conversion streamlines calculations in triple integrals, especially when dealing with geometry involving circular or cylindrical bounds.
Bounded Region
Understanding the bounded region of integration is crucial for setting up a triple integral. In this exercise, this region is denoted by \( E \) which is bounded by specific constraints for each coordinate axis:
  • \(-1 \leq x \leq 1\): This range specifies that our region spans symmetrically along the \( x \)-axis from \(-1\) to \(1\).
  • \(-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\): For each \( x \), the \( y \)-values describe a circular cross-section, centered at the origin, with a radius that depends on \( x \), patterned similarly to a circle of radius one.
  • \(0 \leq z \leq 1-x^2-y^2\): This limits \( z \) from forming anything higher than a paraboloid, creating a dome-shaped cap within the bounded region.
This careful setup allows the integral to describe the volume or other attributes of such regions successfully.
Paraboloid
A paraboloid defines one part of the bounding shape for the solid of integration in this problem. It is represented by the equation \( z = 1 - x^2 - y^2 \), a type of three-dimensional surface that opens downward.When graphed, it shows a curve along each axis and forms a bowl-like shape. Key features of a paraboloid include:
  • The vertex, which is the highest or lowest point, located at \((0,0,1)\) in this case.
  • Cross-sections parallel to the \( xy \)-plane produce circles, indicating its smooth curving nature.
In this exercise, the paraboloid bounds the region vertically, limiting how tall \( z \) can be for each \( x, y \) coordinate pair within the region.
Unit Sphere
The unit sphere plays an important role in determining the outer boundary of our region. In its simplest form, a unit sphere is a sphere with radius 1, centered at the origin, and described by the equation \( x^2 + y^2 + z^2 = 1 \).In relation to our exercise, it effectively 'caps' the paraboloid, ensuring integration does not extend beyond the sphere's limits:
  • For any point \((x, y, z)\) within the region, it must satisfy the spherical boundary condition, which limits all coordinates to within or on the sphere.
  • This defines the maximum reach of \( z \), ensuring it doesn鈥檛 exceed the outside edge of the unit sphere.
Integrating over this shape considers only what's inside, tightly controlled by both the paraboloid and the sphere.

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Most popular questions from this chapter

In the following exercises, the function \(T: S \rightarrow R, T(u, v)=(x, y) \quad\) on the region \(S=\\{(u, v) \mid 0 \leq u \leq 1,0 \leq v \leq 1\\}\) bounded by the unit square is given, where \(R \subset \mathrm{R}^{2}\) is the image of \(S\) under \(T\). a. Justify that the function \(T\) is a \(C^{1}\) transformation. b. Find the images of the vertices of the unit square \(S\) through the function \(T\). c. Determine the image \(R\) of the unit square \(S\) and graph it. $$ x=u^{3}, y=v^{3} $$

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(\quad y\) -axis, and origin, respectively. $$ R \text { is the unit disk; } \rho(x, y)=3 x^{4}+6 x^{2} y^{2}+3 y^{4} \text { . } $$

The power emitted by an antenna has a power density per unit volume given in spherical coordinates by \(p(\rho, \theta, \varphi)=\frac{P_{0}}{\rho^{2}} \cos ^{2} \theta \sin ^{4} \varphi, \quad\) where \(P_{0}\) is a constant with units in watts. The total power within a sphere \(B\) of radius \(r\) meters is defined as \(P=\iiint_{B} p(\rho, \theta, \varphi) d V .\) Find the total power \(P\).

Show that the volume of a regular right hexagonal prism of edge length \(a\) is \(\frac{3 a^{3} \sqrt{3}}{2}\) by using triple integrals.

[T] The average density of a solid \(Q\) is defined as \(\rho_{\text {ave }}=\frac{1}{V(Q)}\) ? \(\rho(x, y, z) d V=\frac{m}{V(Q)},\) where \(V(Q)\) and \(m\) are the volume and the mass of \(Q\), respectively. If the density of the unit ball centered at the origin is \(\rho(x, y, z)=e^{-x^{2}-y^{2}-z^{2}},\) use a CAS to find its average density. Round your answer to three decimal places.
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