Question

Find the area of the part of the cone $z^2=3(x^2+y^2)$ which is inside the sphere $x^2+y^2+z^2=16$.

Short answer

The area is $4\pi \sqrt{7}$.

Step-by-step solution

Understand the Shapes

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

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 $x^2+y^2=4$.

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 $z=\sqrt{3r^2}=\sqrt{12}=2\sqrt{3}$.

Set Up the Surface Integral

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

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, $dS=\frac{\sqrt{7}}{2}rdrd\theta$.

Integrate Over the Region

Integrate $dS$ over the region where $0\le r\le 2$ and $0\le \theta \le 2\pi$: ${\int }_0^{2\pi }{\int }_0^2\frac{\sqrt{7}}{2}rdrd\theta$.

Compute the Integral

$\frac{\sqrt{7}}{2}\cdot {\int }_0^{2\pi }d\theta \cdot {\int }_0^{2}rdr=\frac{\sqrt{7}}{2}\cdot 2\pi \cdot {\left[\frac{r^2}{2}\right]}_0^2=\frac{\sqrt{7}}{2}\cdot 2\pi \cdot \frac{4}{2}=4\pi \sqrt{7}$.

Key concepts

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:
$dS=\sqrt{1+{\left(\frac{\partial z}{\partial r}\right)}^2}\cdot rdrd\theta$.
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.