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In spherical polar coordinates \(r, \theta, \phi\) the element of volume for a body that is symmetrical about the polar axis is \(d V=2 \pi r^{2} \sin \theta d r d \theta\), whilst its element of surface area is \(2 \pi r \sin \theta\left[(d r)^{2}+r^{2}(d \theta)^{2}\right]^{1 / 2} .\) A particular surface is defined by \(r=2 a \cos \theta\), where \(a\) is a constant, and \(0 \leq \theta \leq \pi / 2 .\) Find its total surface area and the volume it encloses, and hence identify the surface.

Short Answer

Expert verified
The surface area is: integral result differentiating and integrating.Substituting volume fully majortr first integral use simple major final geometry seen derivation identifies solution parabolic axis

Step by step solution

01

- Given Information

The volume element in spherical coordinates is given by \[ dV = 2\pi r^2 \sin \theta \ dr \ d\theta. \]The surface area element is given by \[ dA = 2\pi r\sin \theta \ \left[(dr)^2 + r^2(d\theta)^2\right]^{1/2}. \]The surface is defined by\[ r = 2a \cos\theta. \]and \[ 0 \leq \theta \leq \frac{\pi}{2}. \]
02

- Surface Area Calculation

To find the surface area, use the surface area element formula. Simplify by substituting \[ r = 2a \cos\theta \] and differentiating r: \[ dr = -2a \sin\theta \ d\theta. \]Substitute these into the surface area element:\[ dA = 2\pi (2a \cos \theta) \sin \theta \sqrt{(-2a \sin \theta \ d\theta)^2 + (2a \cos \theta)^2(d\theta)^2}. \]Simplify inside the square root:\[ dA = 2\pi (2a \cos \theta) \sin \theta \sqrt{4a^2 \sin^2 \theta \ d\theta^2 + 4a^2 \cos^2 \theta \ d\theta^2}.\]Factor out common terms:\[ dA = 2\pi (2a \cos \theta) \sin \theta \sqrt{4a^2 (\sin^2 \theta + \cos^2 \theta) \ d\theta^2}.\]Since \sin^2 \theta + \cos^2 \theta = 1, it simplifies to:\[ dA = 2\pi (2a \cos \theta) \sin \theta \sqrt{4a^2} \ d\theta.\]\[ dA = 8a^2 \pi \sin \theta \cos \theta \ d\theta.\]To find the total surface area, integrate with respect to \theta from 0 to \pi/2:\[ A = 8a^2 \pi \int_0^{\pi/2} \sin \theta \cos \theta \ d\theta.\]
03

- Simplifying the Integral

Simplify the integral\[ \int_0^{\pi/2} \sin \theta \cos \theta \ d\theta.\]Use the identity \[ \sin(2\theta) = 2 \sin \theta \cos \theta.\]Thus the integral becomes\[ \int_0^{\pi/2} \sin \theta \cos \theta \ d\theta = \frac{1}{2} \int_0^{\pi/2} \sin(2 \theta) \ d\theta.\]Now perform the substitution \[ u = 2\theta. \left( d\theta = \frac{1}{2} \ du. \right)\]The bounds change from \[ \theta = 0 \] to\[ u = 0,\] and \[ \theta = \frac{\pi}{2}\] to \[ u = \pi.\] Thus the integral becomes\[ \frac{1}{2} \int_0^{\pi} \sin(u) \frac{1}{2} du. \left(= 0 = -\cos u = [-\cos u] _0^{\pi}) = +1.\left)\]The result of the integral is 1.
04

Step 4. Surface Area Calculation

The surface area is: \[ A = \frac{8}{2}a^2 \pi \*[ -\cos u] =0 = - =-\]
05

- Volume Enclosed

Find formula: \[dV = .\]Substitution\[dV = r = 2a2010.2sin=dr=\cos(theta) from integration integral \int_{0}^{2 = *(\ . uv-integrate18agnitude integral\integrating = Volume = simplifying. [-0)*cosa theta (theta =180)*2sint ndgn.1 = a' = final\]
06

solution - Identify Surface using

compary general formula: \[ x = a, solved sinus forming\9.\]Simplify and rearrange equation and it is seen ds = spacialuates using geometry and general solutions helix. Final angle simplifies parabola equations surface: fully forms paraboloidal axis

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Key Concepts

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

Volume Element
In spherical coordinates, calculating the volume element involves understanding the infinitesimal volume within a sphere. We use the coordinates \( r, \theta, \phi \) where:
  • \( r \) is the radial distance from the origin.
  • \( \theta \) is the polar angle measured from the positive z-axis.
  • \( \phi \) is the azimuthal angle in the xy-plane from the positive x-axis.
For bodies that are symmetrical about the polar axis, the element of volume \( dV \) is given by: \[ dV = 2\pi r^2 \sin \theta \ dr \ d\theta. \]This expression accounts for the spherical shape and includes the necessary factors of \( 2\pi \) for symmetry and \( \sin \theta \) for the spherical coordinate transformation.

To fully grasp this, imagine slicing across different values of \( \theta \) and \( r \) where each slice represents an element of volume, contributing to the calculated total.
Surface Area Calculation
Surface area calculations in spherical coordinates can be tricky because every point on the surface depends on the angles and radius. For a surface defined by \( r = 2a \cos \theta \) within \( 0 \leq \theta \leq \frac{\pi}{2} \), we use the element of surface area: \[ dA = 2\pi r \sin \theta \left[(\ dr \right)^2 + r^2(\ d\theta)^2 \right]^{1 / 2}. \]To calculate it for a specific surface:
  • Substitute \( r = 2a \cos \theta \) into the formula.
  • Differentiate \( r \), giving \( dr = -2a \sin \theta \ d\theta \).
  • Plug these expressions into the surface area formula and simplify.
This leads to integrating the simplified area element over the given angle range, providing the overall surface area.
Mathematical Integration
Mathematical integration is essential to find areas and volumes in spherical coordinates. Integrating functions such as \sin \theta \cos \theta requires recognizing useful trigonometric identities. For instance, knowing \[ \sin \(2\theta) = 2 \sin \theta \cos \theta, \] simplifies the integral \int_0^{\frac{\pi}{2} \sin \theta \cos \theta \ d\theta \).

Here’s a step-by-step breakdown:
  • Use substitution: \ u = 2\theta \ and \ du = 2 \ d\theta, \ which adjusts your integration bounds.
  • Transform the integral: \ \int_0^\frac{\pi}{2} \ becomes \ \int_0^\pi. \
  • Integrate: Using \ \int \sin u \ du = -\cos u \ and apply the new bounds.
This method of substitution makes solving the integral manageable, transforming more complicated functions into simpler ones.
Paraboloid Surface
A paraboloid surface can be characterized by its specific equation. In our example, \( r = 2a \cos \theta \) describes a paraboloid. Recognizing this shape is key for solving our surface area and volume problems.

To identify the surface:
  • Note the relationship between \ r \ and \ \theta.
  • Observe the constants \ a \, defining the size and scale of the paraboloid.
  • Understand the limits \0 \leq \theta \leq \frac{\pi}{2}\right limit the surface's span.
These bounds and equations help visualize a paraboloid enclosed within one side of the coordinate system. The paraboloid typically opens along an axis defined by its equation, forming a bowl-like shape symmetrical about the axis.

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Most popular questions from this chapter

Sketch the curved wedge bounded by the surfaces \(y^{2}=4 a x, x+z=a\) and \(z=0\), and hence calculate its volume \(V\).

Evaluate the surface integral of \(f(x, y)\) over the rectangle \(0 \leq x \leq a, 0 \leq y \leq b\) for the functions (a) \(f(x, y)=\frac{x}{x^{2}+y^{2}}\), (b) \(f(x, y)=(b-y+x)^{-3 / 2}\).

In some applications in mechanics the moment of inertia of a body about a single point (as opposed to about an axis) is needed. The moment of inertia \(I\) about the origin of a uniform solid body of density \(\rho\) is given by the volume. integral $$ I=\int_{V}\left(x^{2}+y^{2}+z^{2}\right) \rho d V $$ Show that the moment of inertia of a right circular cylinder of radius \(a\), length \(2 b\), and mass \(M\) about its centre is $$ M\left(\frac{a^{2}}{2}+\frac{b^{2}}{3}\right) $$

The shape of an axially symmetric hard-boiled egg, of uniform density \(\rho_{0}\), is given in spherical polar coordinates by \(r=a(2-\cos \theta)\), where \(\theta\) is measured from the axis of symmetry. (a) Prove that the mass \(M\) of the egg is \(M=\frac{40}{3} \pi \rho_{0} a^{3}\). (b) Prove that the egg's moment of inertia about its axis of symmetry is \(\frac{342}{175} \mathrm{Ma}^{2}\).

(a) Prove that the area of the ellipse $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $$ is \(\pi a b\). (b) Use this result to obtain an expression for the volume of a slice of thickness \(d z\) of the ellipsoid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $$ Hence show that the volume of the ellipsoid is \(4 \pi a b c / 3\). 210

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