Find the area of the part of the cone \(z^{2}=3\left(x^{2}+y^{2}\right)\) which is inside the sphere \(x^{2}+y^{2}+z^{2}=16\).

Short Answer

Expert verified
The area is \(4\pi \sqrt{7}\).

Step by step solution

01

Understand the Shapes

The given cone is described by the equation \(z^2 = 3(x^2 + y^2)\) and the sphere is given by \(x^2 + y^2 + z^2 = 16\).
02

Find the Intersection

Substitute the equation of the cone into the sphere's equation: \(x^2 + y^2 + 3(x^2 + y^2) = 16\). This gives \(4(x^2 + y^2) = 16\), so \(x^2 + y^2 = 4\).
03

Rewrite the Equations in Cylindrical Coordinates

In cylindrical coordinates, we use \(r^2 = x^2 + y^2\). So, \(r^2 = 4\) or \(r = 2\). The cone becomes \(z = \sqrt{3r^2} = \sqrt{12} = 2\sqrt{3}\).
04

Set Up the Surface Integral

The surface area element in cylindrical coordinates is \(dS = \sqrt{1 + (\frac{\partial z}{\partial r})^2} \cdot r dr d\theta\). For the cone, \(\frac{\partial z}{\partial r} = \frac{3r}{z} = \frac{3r}{2\sqrt{3}r} = \frac{\sqrt{3}}{2}\). Thus, \(dS = \sqrt{1 + \left( \frac{\sqrt{3}}{2} \right)^2} \cdot r dr d\theta\).
05

Simplify the Surface Integral

\(\sqrt{1 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{1 + \frac{3}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}\). Therefore, \(dS = \frac{\sqrt{7}}{2} r dr d\theta\).
06

Integrate Over the Region

Integrate \(dS\) over the region where \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\): \( \int_0^{2\pi} \int_0^2 \frac{\sqrt{7}}{2} r dr d\theta\).
07

Compute the Integral

\( \frac{\sqrt{7}}{2} \cdot \int_0^{2\pi} d\theta \cdot \int_0^2 r dr = \frac{\sqrt{7}}{2} \cdot 2\pi \cdot \left[ \frac{r^2}{2} \right]_0^2 = \frac{\sqrt{7}}{2} \cdot 2\pi \cdot \frac{4}{2} = 4\pi \sqrt{7}\).

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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 critical tool in multivariable calculus. They are especially useful when dealing with problems that exhibit rotational symmetry around an axis, simplifying calculations greatly.
In cylindrical coordinates, a point in space is described by three coordinates \(r, \theta, z\):
\
    \r - the radial distance from the origin, \theta - the angle around the z-axis, \z - the height above the xy-plane
For instance, the equations of a cone and sphere in our problem become much simpler. We use the substitution \( r^2 = x^2 + y^2 \) resulting in \( z^2 = 3r^2 \) for the cone and \( r^2 + z^2 = 16 \) for the sphere.
Surface Integrals
Surface integrals allow us to integrate over a surface rather than a line or a volume. These are crucial in calculating the surface area of complex shapes like cones or spheres.
To perform surface integrals, we need an expression for the surface area element \(dS\). This element varies based on the coordinate system and the surface itself.
In cylindrical coordinates, for a surface \( z = f(r, \theta) \), the surface area element is given by:
\( dS = \sqrt{1 + \left( \frac{\partial z}{\partial r} \right)^2} \cdot r \ dr \ d\theta \).
This expression takes into account the slope of the surface and the area in the xy-plane to ensure the integration covers the actual surface.
Geometry of Cones and Spheres
Understanding the geometric properties of cones and spheres is foundational to solving surface area problems in 3D space.
A cone has a circular base and tapers smoothly from the base to a point called the vertex. If described by the equation \(z^2 = 3(x^2 + y^2)\) in Cartesian coordinates, it converts to \( z = \sqrt{3} r \) in cylindrical coordinates.
On the other hand, a sphere is a perfectly symmetrical 3D shape where every point on the surface is equidistant from the center. The sphere’s Cartesian equation \( x^2 + y^2 + z^2 = 16 \) becomes \( r^2 + z^2 = 16 \) in cylindrical coordinates.
The intersection of these two shapes provides insights into setting up and computing our integrals.
Change of Variables
Changing variables is a technique that simplifies the problem by converting a challenging set of coordinates into an easier one. In our case, changing from Cartesian to cylindrical coordinates lets us leverage symmetrical properties.
In general, a change of variables involves substituting new variables into the problem and adjusting all expressions and limits accordingly. For instance, if \( x = r \cos\theta, y = r \sin\theta, \) we can replace \( x \) and \( y \) in the equations. This step can simplify the integration process significantly, often converting it into more manageable bounds and shapes.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus into more dimensions, allowing us to explore and solve more complex problems.
It covers differentiation and integration in higher dimensions, factoring in partial derivatives, gradients, and multiple integrals.
Dealing with surface area calculation is one application among many in multivariable calculus. To solve these problems, we often use concepts like surface integrals, gradient fields, and transformations between different coordinate systems.
This enhances our ability to analyze and compute properties of intricate geometrical shapes in higher-dimensional spaces.

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Most popular questions from this chapter

Find the volume between the planes \(z=2 x+3 y+6\) and \(z=2 x+7 y+8,\) and over the square in the \((x, y)\) plane with vertices (0,0),(1,0),(0,1),(1,1).

As needed, use a computer to plot graphs and to check values of integrals. (a) Find the centroid of the area between the \(x\) axis and one arch of \(y=\sin x\). (b) Find the volume formed if the area in (a) is rotated about the \(x\) axis. (c) Find \(I_{x}\) of a mass of constant density occupying the volume in (b).

The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). Prove the "parallel axis theorem": The moment of inertia I of a body about a given axis is \(I=I_{m}+M d^{2},\) where \(M\) is the mass of the body, \(I_{m}\) is the moment of inertia of the body about an axis through the center of mass and parallel to the given axis, and \(d\) is the distance between the two axes.

Verify each of the following answers for an indefinite integral by one or more of the methods suggested above. \(\int \frac{K d r}{\sqrt{1-K^{2} r^{2}}}=\sin ^{-1} K r \quad\) or \(\quad-\cos ^{-1} K r \quad\) or \(\quad \tan ^{-1} \frac{K r}{\sqrt{1-K^{2} r^{2}}}.\) Hints: Sketch a right triangle with acute angles \(u\) and \(v\) and label the sides so that \(\sin u=K r .\) Also note that \(u+v=\pi / 2 ;\) then if \(u\) is an indefinite integral, so is \(-v\) since they differ by a constant of integration.

The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). A rectangular lamina has vertices (0,0),(0,2),(3,0),(3,2) and density \(x y .\) Find (a) \(M\), (b) \(\bar{x}, \bar{y}\), (c) \(I_{x}, I_{y}\), (d) \( I_{m}\) about an axis parallel to the \(z\) axis. Hint: Use the parallel axis theorem and the perpendicular axis theorem.

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