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Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{\pi / 6}^{\pi / 3} \int_{0}^{2 \csc \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Short Answer

Expert verified
Question: Find the value of the following integral in spherical coordinates: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta$$ Answer: The value of the integral is: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta = \frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$

Step by step solution

01

Transform the integral

We need to transform the Cartesian integral to spherical coordinates with the expression for volume in spherical coordinates, using the Jacobian for the transformation: $$\int_{0}^{2 \pi} \int_{\pi / 6}^{\pi / 3} \int_{0}^{2 \csc \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$
02

Integrate with respect to ρ

Evaluate the innermost integral with respect to radial distance \(\rho\): $$\int_{0}^{2 \csc \varphi} \rho^2 \sin \varphi d \rho = \left[\frac{1}{3}\rho^3\sin\varphi\right]\bigg|_0^{2\csc\varphi} $$ This results in: $$\left[\frac{1}{3}(2\csc\varphi)^3\sin\varphi\right] = \frac{8}{3}\sin\varphi$$
03

Integrate with respect to 𝜑

Now, we'll evaluate the middle integral with respect to polar angle \(\varphi\): $$\int_{\pi/6}^{\pi/3} \frac{8}{3}\sin\varphi d\varphi = \left[-\frac{8}{3}\cos\varphi\right]\bigg|_{\pi/6}^{\pi/3} $$ This results in: $$-8\left(-\frac{1}{3} - \frac{\sqrt{3}}{6}\right) = \frac{8}{3}+\frac{4\sqrt{3}}{3}$$
04

Integrate with respect to 𝜃

Finally, evaluate the outermost integral with respect to azimuthal angle θ: $$\int_{0}^{2\pi}\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)d\theta=\left[\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)\theta\right]\bigg|_{0}^{2\pi}$$ This results in: $$\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)(2\pi)=\frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$
05

Final Result

The value of the integral is: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta = \frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$

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Key Concepts

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

Triple Integral
The triple integral is a powerful mathematical tool that extends the concept of single and double integrals to three dimensions. It allows us to calculate the volume under a surface in three-dimensional space and is particularly useful in physics and engineering for solving problems involving mass, density, and volume. A triple integral is represented by the notation \( \int \int \int f(x, y, z) \, dx \, dy \, dz \), where \( f(x, y, z) \) is a three-variable function, and \( dxdydz \) signifies the infinitesimal volume elements. To perform a triple integral, we usually integrate in the order from the innermost to outermost integral, while the limits of integration represent the region we're integrating over.

When dealing with regions that are not easily describable in Cartesian coordinates, such as spheres and cones, it becomes more efficient to use a coordinate system that matches the symmetry of the problem, like spherical coordinates.
Spherical Integration
Spherical integration is particularly useful when dealing with problems that have spherical symmetry. In such cases, spherical coordinates (\(\rho, \varphi, \theta\)) are employed, where \(\rho\) is the distance from the origin, \(\varphi\) is the polar angle measured from the positive z-axis, and \(\theta\) is the azimuthal angle in the xy-plane from the positive x-axis. Spherical integrals involve integrating over these three variables, accounting for the spherical volume element \( dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \).

To compute a volume in spherical coordinates, one typically evaluates an integral of this form: \( \int_{\theta_1}^{\theta_2} \int_{\varphi_1}^{\varphi_2} \int_{\rho_1}^{\rho_2} f(\rho, \varphi, \theta) \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \). The limits of integration \(\theta_1, \theta_2, \varphi_1, \varphi_2, \rho_1, \rho_2\) define the spherical region of integration.
Volume in Spherical Coordinates
Calculating the volume in spherical coordinates involves using the spherical volume element \( dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \), which arises from the geometry of the spherical coordinate system. The volume element represents a small 'chunk' of space, corresponding to a wedge-shaped segment of a sphere.

To find the volume of a spherical region, we integrate the volume element over the desired region. For example, the formula for the volume of a sphere of radius R is obtained by integrating the volume element from \(0\) to \(R\) for \(\rho\), from \(0\) to \(\pi\) for \(\varphi\), and from \(0\) to \(2\pi\) for \(\theta\): \( V = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \). This results in the familiar \(\frac{4}{3}\pi R^3\), the volume of a sphere.
Jacobian
The Jacobian is a determinant that arises in multivariable calculus, specifically in transformations between coordinate systems. It's used to scale the volume when changing from one coordinate system to another, such as from Cartesian to spherical coordinates. Mathematically, the Jacobian for spherical coordinates \(J(\rho, \varphi, \theta)\) is \( \rho^2 \sin\varphi \), and it represents how much the volume element \(dV\) changes when transforming between these systems.

The Jacobian is crucial because it ensures the preservation of the proper scale of volume, area, or length during integration. For instance, when computing integrals in spherical coordinates, we must multiply the integrand by the Jacobian to account for the distortion in volume elements due to the coordinate change. This is precisely why the volume element in spherical coordinates includes the Jacobian factor \( \rho^2 \sin\varphi \).

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Most popular questions from this chapter

The following table gives the density (in units of \(\mathrm{g} / \mathrm{cm}^{2}\) ) at selected points of a thin semicircular plate of radius 3. Estimate the mass of the plate and explain your method. $$\begin{array}{|c|c|c|c|c|c|} \hline & \boldsymbol{\theta}=\mathbf{0} & \boldsymbol{\theta}=\boldsymbol{\pi} / \boldsymbol{4} & \boldsymbol{\theta}=\boldsymbol{\pi} / \boldsymbol{2} & \boldsymbol{\theta}=\boldsymbol{3} \pi / \boldsymbol{4} & \boldsymbol{\theta}=\boldsymbol{\pi} \\ \hline \boldsymbol{r}=\mathbf{1} & 2.0 & 2.1 & 2.2 & 2.3 & 2.4 \\ \hline \boldsymbol{r}=\mathbf{2} & 2.5 & 2.7 & 2.9 & 3.1 & 3.3 \\ \hline \boldsymbol{r}=\mathbf{3} & 3.2 & 3.4 & 3.5 & 3.6 & 3.7 \\ \hline \end{array}$$

Density distribution A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3}\)

Use spherical coordinates to find the volume of the following solids. A ball of radius \(a>0\)

General volume formulas Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Frustum of a cone Find the volume of a truncated solid cone of height \(h\) whose ends have radii \(r\) and \(R\).

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