Caluclus: Table of contents
Vector Fields. Parametric Equations of Curves and Surfaces
Vector fields
A function whose codomain is \(\mathbb R^2\) or \(\mathbb R^3\) is called a vector field. This name emphasize that the output of the function is a vector. Most vector functions that we will consider will have a domain that is a subset of \(\mathbb R\), \(\mathbb R^2\), or \(\mathbb R^3\).
An example of a vector field is the function \(\overrightarrow{F}:[0,5]\to\mathbb R\) defined as \[\overrightarrow F(t)=\langle 7t+5, t^2+4\rangle.\]
Parametric equations of curves
We will now see how vector fields whose domains are subsets of \(\mathbb R\) correspond to curves. Consider a vector field \(\overrightarrow{r}:[0,2\pi)\to\mathbb R^2\) defined as: \(\overrightarrow{r}(t)=\langle 3\cos t,3\sin t\rangle\). Each number \(t\in[0,2\pi)\) is mapped to a vector \(\langle 3\cos t,3\sin t\rangle\). Let us place this vector so its starting point is at \((0,0)\), and let us put green paint at the tip of the vector. Then as \(t\) ranges from \(0\) to \(2\pi\) the tip of the vector will leave a green trace. Since \((3\sin t)^2+(3\cos t)^2=1\), the tip of the vector will trace a green circle in the plane. The equation \(\overrightarrow r(t)=\langle 3\cos t, 3 \sin t\rangle\), \(t\in[0,2\pi)\) is called the parametric equations of the circle. We use plural here, because parametric equations are often written in the following way:
\begin{eqnarray*}
x(t)&=&3\cos t\newline
y(t)&=&3\sin t\newline
0\leq&t&< 2\pi.
\end{eqnarray*}
This way we can assign a curve to every vector field \(\overrightarrow r:[a,b]\to\mathbb R^2\) or \(\overrightarrow r:[a,b]\to\mathbb R^2\). The curve is obtained as a trace of the tip of the vector whose starting point is at the origin.
The formal definition is: A curve corresponding to the vector field \(\overrightarrow{r}:[a,b]\to\mathbb R^3\) is the image of the mapping \(t\mapsto (0,0,0)+\overrightarrow{r}(t)\).
Parametric equations of surfaces
In an analogy to the previous section, we can assign a surface to every vector field of the form \(\overrightarrow r: D\to\mathbb R^3\), where \(D\subseteq \mathbb R^2\) is a two-dimensional region.
Consider, for example, the following vector field: \(\overrightarrow r:[0,2\pi)\times[0,\pi]\to \mathbb R^2\) defined by \[\overrightarrow r(\theta,\phi)=\langle 3\cos\theta\sin\phi,3 \sin\theta\sin \phi,3\cos \phi\rangle.\] When \(\theta\) and \(\phi\) take all possible values in \([0,2\pi)\times[0,\pi]\) the tip of the vector \(\overrightarrow r\) will trace the sphere of radius \(3\). It is common to write these parametric equations in the form:
\begin{eqnarray*}
x(\theta,\phi)&=&3\cos\theta\sin\phi\\
y(\theta,\phi)&=&3\sin\theta\sin\phi\\
z(\theta,\phi)&=&3\cos\phi\\
0\leq&\theta&< 2\pi\\
0\leq&\phi&\leq\pi.
\end{eqnarray*}
The number \(\phi\) is the angle between the vector \(\overrightarrow{r}(\theta, \phi)\) and the \(z\)-axis, while the number \(\theta\) is the angle between the \(x\)-axis and the projection of \(\overrightarrow r(\theta, \phi)\) to the \(xy\)-plane.
If we change the domain of the previous vector field to \([0,2\pi)\times \left[\frac{\pi}3,\frac{2\pi}3\right]\) then we obtain the sphere with two pieces removed (one near the north pole and one near the south pole). The picture below is obtained using the program Mathematica (wolfram.com) with the command:
ParametricPlot3D[{3 Cos[t] Sin[s], 3 Sin[t] Sin[s], 3 Cos[s]}, {t, 0, 2 Pi}, {s, Pi/3, 2 Pi/3}]
You may visit wolframalpha.com and type the previous command to obtain the same figure. By changing the range for \(t\) and \(s\) you can get different pieces of the sphere.
Example
Find the parametric equations of the torus obtained by rotating the circle \((y-3)^2+z^2=1\) in the \(yz\)-plane around the \(z\)-axis.
The parametric equations of the circle in the \(yz\) can be written in the vector form:
\[\overrightarrow r(t)=\langle 0, 3+\cos t, \sin t\rangle,\]
\(0\leq t\leq 2\pi\). The equation of the circle obtained by rotating the point \((0,y_0,z_0)\) around \(z\) axis is: \(\langle y_0\sin s, y_0\cos s,z_0\rangle\), \(0\leq s\leq 2\pi\). Hence the equation of the torus is:
\begin{eqnarray*}x(t,s)&=&(3+\cos t)\sin s\\
y(t,s)&=& (3+\cos t)\cos s\\
z(t,s)&=&\sin t\\
0\leq &t&\leq 2\pi\\
0\leq&s&\leq 2\pi.
\end{eqnarray*}
You can generate the following picture by pasting the command:
ParametricPlot3D[{(3 + Cos[t]) Sin[s], (3 + Cos[t]) Cos[s], Sin[t]}, {t, 0, 2 Pi}, {s, 0, 2 Pi}]
in
wolframalpha.com.
Example
Let \(L_1\) be the line obtained as the intersection of the planes \(3x-y-z=3\) and \(x+y-z=1\). Let \(L_2\) be the line given by \(\langle -1,0,1\rangle+t\langle 2,1,1\rangle\).
- (a) Find the parametric equations of the line \(L_1\).
- (b) Prove that the lines \(L_1\) and \(L_2\) are skew lines.
- (a) The parametric equations of \(L_1\) are obtained by solving the system \begin{eqnarray*}
3x-y-z&=&3\\ x+y-z&=&1.
\end{eqnarray*}
Adding the two equations yields \(4x-2z=4\) and if we choose \(x=t\) we get \(z=2t-2\) and \(y=1+z-x=
1+2t-2-t=t-1\). Hence the parametric equations of \(L_1\) are:
\begin{eqnarray*}
x&=&t\\
y&=&t-1\\
z&=&2t-2.
\end{eqnarray*}
- (b) The parametric equations of \(L_2\) can be written as \begin{eqnarray*}
x&=&-1+2s\\
y&=&s\\
z&=&1+s.
\end{eqnarray*}
If the lines \(L_1\) and \(L_2\) intersect, then the system of equations \begin{eqnarray*}
t&=&-1+2s\\
t-1&=&s\\
2t-2&=&1+s
\end{eqnarray*} would have a solution. However, from the first two equations we get
\(s=2\) and \(t=3\), which does not satisfy \(2t-2=1+s\). Since the lines are not parallel (\(\langle 1,1,2\rangle\) is not parallel to \(\langle 2,1,1\rangle\)), they are skew.
Example
Find parametric equations of the curve that is obtained as the intersection of the paraboloid \(z=9x^2+4y^2\) and the cylinder \(x^2+y^2=16\).
The parametric equations of the cylinder are \(\langle x,y,z\rangle=\langle 4\cos\theta, 4\sin\theta, z\rangle\), \(0\leq \theta < 2\pi\), \(-\infty < z < +\infty\). Points of the cylinder that belong to the paraboloid must satisfy: \begin{eqnarray*}
x&=&4\cos\theta\\
y&=&4\sin\theta\\
z&=&144\cos^2\theta+64\sin^2\theta.\end{eqnarray*}
Example
Initially, a paint nozzle is placed at the point \((100,0,0)\) and the nozzle is moving along the line parallel to the \(z\)-axis. The position of the nozzle at time \(t\) is \((100,0,t^2)\).
The paraboloid \(z=x^2+y^2\) is rotating around \(z\)-axis with constant angular speed. As time goes from \(0\) to \(1\), the paraboloid completes one rotation around the \(z\)-axis and ends up in the original position.
If the nozzle is spraying the paint towards the paraboloid the entire time, set up the integral (but do not evaluate) for the length of the curve that is painted on the surface during the time interval from \(t=0\) to \(t=1\). There is no gravity nor air resistance that could interfere with the painting.
The length of the curve is the same as the length of the curve obtained by nozzle that rotates around the paraboloid in unit time. The equation of the rotating nozzle is:
\begin{eqnarray*}
x(t)&=&100\cos (2\pi t)\\
y(t)&=&100\sin (2\pi t)\\
z(t)&=&t^2\\
0\leq&t&\leq 1.
\end{eqnarray*}
When the cylindrical coordinates of the nozzle are \((\theta,r,z)\) the paint on paraboloid will appear at the point whose cylindrical coordinates are \((\theta, \sqrt{z}, z)\). Hence, the parametric equation of the curve is:
\begin{eqnarray*}
x(t)&=&t\cos (2\pi t) \\
y(t)&=&t\sin (2\pi t)\\
z(t)&=&t^2\\
0\leq&t&\leq 1.
\end{eqnarray*}
The velocity vector is \(\vec{v}(t)=\langle \cos (2\pi t)-2\pi t\sin(2\pi t), \sin(2\pi t)+2\pi t\cos(2\pi t), 2t\rangle\), and the speed is \[v(t)=\sqrt{1+4t^2+4\pi^2t^2}=\sqrt{1+\left(4+4\pi^2\right)t^2}.\]
The required length is \begin{eqnarray*}
l&=&\int_0^1\sqrt{1+\left(4+4\pi^2\right)t^2}\,dt.
\end{eqnarray*}
Practice problems
Problem 1. Determine the parametric equations that corresponding to the following surface:
- (A) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r^2\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (B) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5\sin r\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (C) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5(2+\cos r)\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (D) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=\sqrt{9-r^2}\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (E) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
The parametrization A corresponds to a paraboloid that is elongated along \(x\) axis.
The parametrization B corresponds to a surface that is radially periodic. The surface intersects the \(z\) axis at a point \((0,0,0)\), and increasing \(r\) results in the increase of the \(z\)-coordinate.
The parametrization C corresponds to a surface that is radially periodic. The surface intersects the \(z\) axis at a point \((0,0,10)\), and as \(r\) increases, in the neighborhood of the \(z\) axes the height of the surface decreases.
The parametrization D corresponds to a hemisphere because \(z^2=9-x^2-y^2\), or, equivalently \(x^2+y^2+z^2=9\).
The parametrization E corresponds to a cone.
The correct answer is C.
Problem 2. Determine the parametric equations that corresponding to the following surface:
- (A) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r^2\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (B) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5\sin r\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (C) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5(2+\cos r)\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (D) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=\sqrt{9-r^2}\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (E) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
Problem 3. Determine the parametric equations that corresponding to the following surface:
- (A) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r^2\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (B) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5\sin r\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (C) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5(2+\cos r)\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (D) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=\sqrt{9-r^2}\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (E) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
Problem 4. Determine the parametric equations that corresponding to the following surface:
- (A) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r^2\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (B) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5\sin r\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (C) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5(2+\cos r)\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (D) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=\sqrt{9-r^2}\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (E) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
Problem 5. Determine the parametric equations that corresponding to the following surface:
- (A) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r^2\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (B) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5\sin r\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (C) \(\displaystyle x=2r\cos \theta\), \(\displaystyle y=2r\sin \theta\), \(\displaystyle z=5(2+\cos r)\), \(\displaystyle 0\leq r\leq 6\pi\), \(\displaystyle 0\leq \theta < 2\pi\)
- (D) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=\sqrt{9-r^2}\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
- (E) \(\displaystyle x=r\cos \theta\), \(\displaystyle y=r\sin \theta\), \(\displaystyle z=9-r\), \(\displaystyle 0\leq r\leq 3\), \(\displaystyle 0\leq \theta < 2\pi\)
Problem 6. The surface \(S\) is formed by rotating the graph of the function \(y=x^4\) around the \(y\) axis. Which of the following parametrizations is a correct parametrization for \(S\)?
- (A) \(x=t\), \(y=t^2+r^2+(t^2+r^2)^2\), \(z=r\)
- (B) \(x=t\), \(y=(t^2+r^2)^2\), \(z=r\)
- (C) \(x=t\), \(y=\sqrt{t^2+r^2}^3\), \(z=r\)
- (D) \(x=t\), \(y=(t^2+r^2)^3\), \(z=r\)
- (E) \(x=t\), \(y=\sqrt{t^2+r^2}^3+(t^2+r^2)^3\), \(z=r\)
Let us denote \(f(x)=x^4\).
Assume that \(P(t,s,r)\) is an arbitrary point on the surface \(S\). Let us denote by \(d(P,y)\) its distance from the \(y\)-axis.
There is a point \(\tilde P\) in the \(xy\) plane that belongs to the graph of \(y=f(x)=x^4\) and that can be obtained by rotating \(P\) around \(y\)-axis. We have that the \(z\)-coordinate of \(\tilde P\) is \(0\) while its \(y\)-coordinate is the same as the \(y\)-coordinate of \(P\). Let us denote by \(\tilde t\) its \(x\)-coordinate. Therefore we have that the coordinates of \(\tilde P\) are \(\tilde P(\tilde t, s, 0)\).
We have that \(d(\tilde P, y)=d(P,y)\) because the rotation around \(y\) preserves the distance from \(y\). Clearly, \(d(P,y)=\sqrt{t^2+r^2}\) and \(d(\tilde P,y)=\tilde t\), implying \[\tilde t=\sqrt{t^2+r^2}.\] The point \(\tilde P\) is on the graph of \(y=f(x)\) hence \[s=f(\tilde t)= f\left(\sqrt{t^2+r^2}\right)=(t^2+r^2)^2.\]
Thus the parametrization is \(x= t\), \(z=r\), \(y=(t^2+r^2)^2\).
Thus, the correct answer is B.
Problem 7. The surface \(S\) is formed by rotating the graph of the function \(y=x^3+x^6\) around the \(y\) axis. Which of the following parametrizations is a correct parametrization for \(S\)?
- (A) \(x=t\), \(y=t^2+r^2+(t^2+r^2)^2\), \(z=r\)
- (B) \(x=t\), \(y=(t^2+r^2)^2\), \(z=r\)
- (C) \(x=t\), \(y=\sqrt{t^2+r^2}^3\), \(z=r\)
- (D) \(x=t\), \(y=(t^2+r^2)^3\), \(z=r\)
- (E) \(x=t\), \(y=\sqrt{t^2+r^2}^3+(t^2+r^2)^3\), \(z=r\)
Let us denote \(f(x)=x^3+x^6\).
Assume that \(P(t,s,r)\) is an arbitrary point on the surface \(S\). Let us denote by \(d(P,y)\) its distance from the \(y\)-axis.
There is a point \(\tilde P\) in the \(xy\) plane that belongs to the graph of \(y=f(x)=x^3+x^6\) and that can be obtained by rotating \(P\) around \(y\)-axis. We have that the \(z\)-coordinate of \(\tilde P\) is \(0\) while its \(y\)-coordinate is the same as the \(y\)-coordinate of \(P\). Let us denote by \(\tilde t\) its \(x\)-coordinate. Therefore we have that the coordinates of \(\tilde P\) are \(\tilde P(\tilde t, s, 0)\).
We have that \(d(\tilde P, y)=d(P,y)\) because the rotation around \(y\) preserves the distance from \(y\). Clearly, \(d(P,y)=\sqrt{t^2+r^2}\) and \(d(\tilde P,y)=\tilde t\), implying \[\tilde t=\sqrt{t^2+r^2}.\] The point \(\tilde P\) is on the graph of \(y=f(x)\) hence \[s=f(\tilde t)= f\left(\sqrt{t^2+r^2}\right)=\sqrt{t^2+r^2}^3+(t^2+r^2)^3.\]
Thus the parametrization is \(x= t\), \(z=r\), \(y=\sqrt{t^2+r^2}^3+(t^2+r^2)^3\).
Thus, the correct answer is E.
Problem 8. If \(\vec F(x,y,z)=\langle -5x^2, -y, x\rangle\) and \(\vec G(x,y,z)=\langle 3z,z-y, x+y\rangle\) are two vector fields, determine the vector field \(\vec H(x,y,z)=\vec F(x,y,z)\times \vec G(x,y,z)\).
The cross product of the vector fields \(\vec F\) and \(\vec G\) is:
\begin{eqnarray*}
\vec H(x,y,z)&=&\left|\begin{array}{ccc}\vec i & \vec j& \vec k\\
-5x^2 & -y & x\\
3z & z-y & x+y \end{array}\right| = \left(-y\cdot (x+y)-x\cdot (z-y)\right)\vec i -
\left(-5x^2\cdot (x+y)-x\cdot 3z\right)\vec j\\ &&+ \left( -5x^2\cdot (z-y)-(x-y)\cdot 3z\right)\vec k
\\
&=&\langle-y^2-xz, 5x^3+5x^2y+3xz, -5x^2z+5x^2y+3yz\rangle.
\end{eqnarray*}
Problem 9. Determine the parametrization of the curve \(\gamma\) that is the intersection of the paraboloid \(z=16x^2+9y^2\) and the hyperboloid \(z=64+x^2-y^2\).
Let us start from the following parametrization of the paraboloid:
\begin{eqnarray*}
x&=&\frac{r\cos\theta}{4}\\
y&=&\frac{r\sin\theta}{3}\\
z&=&r^2\\
0\leq&\theta&< 2\pi\\
0\leq&r&< +\infty.
\end{eqnarray*}
Each point of the paraboloid, except for the origin, corresponds to a unique choice of \((r,\theta)\in (0,+\infty)\times[0,2\pi)\). In order for \((r,\theta)\) to correspond to a point on the hyperboloid, we must have \[r^2=64+r^2\cdot\left(\frac{\cos^2\theta}{16}-\frac{\sin^2\theta}{9}\right)\] which yields \begin{eqnarray*}
r^2&=&\frac{64}{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16}},\mbox{ and }\\
r&=&\frac{8}{\sqrt{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16} } }.
\end{eqnarray*}
Therefore, the coordinates \((x,y,z)\) of the points at the intersection of two surfaces must satisfy:
\begin{eqnarray*}
x&=& \frac{8\cos\theta}{4\sqrt{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16} } }\\
y&=& \frac{8\sin\theta}{3\sqrt{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16} } }\\
z&=&\frac{64}{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16}}\\
&&\mbox{ for some } \theta \in [0,2\pi).
\end{eqnarray*}
Thus one parametrization is:
\begin{eqnarray*} x&=&\frac{8\cos\theta}{4\sqrt{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16} } },\\
y&=&\frac{8\sin\theta}{3\sqrt{1+\frac{\sin^2\theta}{9}-
\frac{\cos^2\theta}{16} } },\\
z &=&\frac{64}{1+\frac{\sin^2\theta}{9}-\frac{\cos^2\theta}{16}},\\
&&0\leq \theta < 2\pi.
\end{eqnarray*}
Problem 10. Consider the ellipsoid
\[\frac{x^2}{25}+\frac{(y+5)^2}{4}+\frac{z^2}{9}=1.\]
Which of the following expressions is a correct parametrization for the ellipsoid?
- (A) \(x=4\cos\theta\sin\phi\),
\(y=-5+2\sin\theta\sin\phi\),
\(z=5\cos\phi\),
\(0\leq \theta < 2\pi\),
\(0\leq \phi\leq \pi\).
- (B) \(x=16\cos\theta\sin\phi\),
\(y=-5+4\sin\theta\sin\phi\),
\(z=25\cos\phi\),
\(0\leq \theta\leq \pi\),
\(0\leq \phi < 2\pi\).
- (C) \(x=5\cos\theta\sin\phi\),
\(y=-5+2\sin\theta\sin\phi\),
\(z=3\cos\phi\),
\(0\leq \theta < 2\pi\),
\(0\leq \phi\leq \pi\).
- (D)
\(x=16\cos\theta\sin\phi\),
\(y=-5+4\sin\theta\sin\phi\),
\(z=25\cos\phi\),
\(0\leq \theta\leq \pi\),
\(0\leq \phi\leq \pi\).
- (E)
\(x=\cos\theta\sin\phi\),
\(y=-3+\sin\theta\sin\phi\),
\(z=\cos\phi\),
\(0\leq \theta < 2\pi\),
\(0\leq \phi\leq \pi\).
The parametrization for the ellipsoid is:
\begin{eqnarray*}
x&=&5\cos\theta\sin\phi\\
y&=&-5_{e}+2\sin\theta\sin\phi\\
z&=&3\cos\phi\\
0\leq&\theta&< 2\pi\\
0\leq&\phi&\leq\pi.
\end{eqnarray*}
Thus, the correct answer is C.