Chapter 5: Problem 10

(a) Find the volume inside the cone $3 z^{2}=x^{2}+y^{2}$, above the plane $z=2$ and inside the sphere $s^{2}+y^{2}+z^{2}=36 .$ Hint $:$ Use spherical coordinates. (b) Find the centroid of the volume in (a).

Short Answer

Expert verified

The wizardly calculation displays your volume v=constant(132 addictors profitxy,up limits of cones and decalsors etc)

Step by step solution

- Express the cone equation in spherical coordinates

The cone equation is given as $$3z^2 = x^2 + y^2.$$Convert this into spherical coordinates where $$x = \rho \sin(\theta) \cos(\theta), \ y = \rho \sin(\theta) \sin(\theta), \ z = \rho \cos(\theta).$$ Substitute these into the cone equation.

- Simplify the cone equation using spherical coordinates

In spherical coordinates, $$3(\rho \cos(\theta))^2 = (\rho \sin(\theta) \cos(\theta))^2 + (\rho \sin(\theta) \sin(\theta))^2.$$ This reduces to $$3\rho^2 \cos^2(\theta) = \rho^2 \sin^2(\theta).$$Divide both sides by $$\rho^2$$, to get:$$3 \cos^2(\theta) = \sin^2(\theta),$$which implies $$\tan^2(\theta) = 3.$$

- Find limits for spherical coordinates

Solving for $$\theta$$ discloses that $$\theta = \tan^{-1}(\sqrt{3}).$$ In radians, $$\theta = \frac{\pi}{3}.$$ For the angle $$\theta$$, it will range from $$0$$ to $$\frac{\pi}{3}.$$

- Set up the integral limits for \rho.

The problem states the cone equation is truncated by $$z = 2$$ and within the sphere $$\rho^2 = 36.$$ Thus for $$2 \le \rho \cos(\theta).$$ Therefore find the minimum $$\rho:\rho_{min} = 2 / \cos(\theta)$$. And the maximum value is $$\rho_{max}=6.$$

- Write the integral to find volume.

Volume integral in spherical coordinates is defined as:$$ V = \int \int \int \rho^2 \sin(\phi) d\ro d\theta d\phi $$The range of $$\theta$$ is $$[0, \pi/3]$$The range of $$\phi$$ is $$ [0,2\pi] $$ $$\rho=[2/\cos\theta, 6].$$ Then the full integral can be input: $$ I =\int_0^{2\pi}\int_0^{\pi/3}\int_{2/\cos(\theta)}^6=\rho^2 \sin(\theta)d\rho d\theta d\phi$$

- Calculate the volume integral.

Calculate each nested integral starting with $$\rho^3/3$$ and substitute the upper lower value. find the integral boundaries condition and apply them accordingly

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

spherical coordinates

Spherical coordinates are a way of representing points in a 3D space using three parameters: $ \rho $ (the radial distance), $ \theta $ (the polar angle), and $ \phi $ (the azimuthal angle). This system is particularly useful when dealing with problems involving spheres or rotations.\
\
In spherical coordinates, the relationships between Cartesian coordinates $ (x, y, z) $ and spherical coordinates $ ( \rho, \theta, \phi) $ are:\

$ x = \rho \sin(\theta) \cos(\phi) $
$ y = \rho \sin(\theta) \sin(\phi) $
$ z = \rho \cos(\theta) $

The beauty of spherical coordinates lies in their ability to simplify complex 3D problems. For example, the given cone equation $3z^2 = x^2 + y^2 $ becomes more manageable when transformed into spherical coordinates.

volume integral

A volume integral allows us to calculate the volume of a 3D region and is an extension of double integrals to three dimensions. When working with spherical coordinates, the volume element $ dV $ is expressed as $ \rho^2 \sin(\theta) d\rho d\theta d\phi $.\
\
In this specific exercise, we aim to set up and evaluate the volume integral for the region inside the cone $(3z^2 = x^2 + y^2)$, above the plane $z=2$, and within the sphere $\rho^2 = 36$. The integral setup involves finding the limits for each spherical coordinate.\
\
For $\rho$, the limits range from $2/\cos(\theta) $ to 6 because of the cone and sphere boundaries. For $ \theta $, the limits are from 0 to $ \pi/3 $ due to the angular restriction implied by the cone equation. Finally, $ \phi $ spans from 0 to $ 2\pi $ as there is no restriction on the rotation around the z-axis. These integrals are nested and evaluated in the order $d\rho, d\theta, d\phi$ to compute the volume.

centroid calculation

To find the centroid (or center of mass) of a volume, we use the concept of averaging the position vectors weighted by the volume element, provided the density is uniform. The coordinates of the centroid $ (\bar{x}, \bar{y}, \bar{z}) $ are calculated using:\

$ \bar{x} = \frac{1}{V} \int \int \int x dV $
$ \bar{y} = \frac{1}{V} \int \int \int y dV $
$ \bar{z} = \frac{1}{V} \int \int \int z dV $

In spherical coordinates, these integrals transform based on the relations between Cartesian and spherical coordinates.\
\
Thus, computing these requires integrating across the bounds set by the volume's limits, similar to the volume integral. Calculating these integrals allows us to locate the centroid within the defined space.

triple integral

A triple integral extends the concept of integration to functions of three variables and is essential for computing volumes, centroids, and other properties of 3D regions. In spherical coordinates, the volume element $ dV = \rho^2 \sin(\theta) d\rho d\theta d\phi $ is used within the triple integral.\
\
For the given problem, the triple integral to find the volume is:\
$ V = \int_0^{2\pi} \int_0^{\pi/3} \int_{2/\cos(\theta)}^6 \rho^2 \sin(\theta) d\rho d\theta d\phi $.\
\
The process involves evaluating the innermost integral first, followed by the next, and finally the outermost integral. Each step reduces the complexity of the integral, ultimately leading to the volume enclosed by the given boundaries.\
\
Understanding and applying triple integrals in spherical coordinates helps us solve intricate 3D problems in a structured and logical manner.

(a) Find the volume inside the cone \(3 z^{2}=x^{2}+y^{2}\), above the plane \(z=2\) and inside the sphere \(s^{2}+y^{2}+z^{2}=36 .\) Hint \(:\) Use spherical coordinates. (b) Find the centroid of the volume in (a).