Math 230-2: Kursobjekt

Looking at the equation, we can take \(\boldn=\angvec{1,2,-3}\) to be our normal vector. Next, the point \(P=(5,0,0)\) is easily seen to be a point of \(\mathcal{S}\text{.}\) To find vectors \(\boldw=\angvec{w_1,w_2,w_3}\) orthogonal to \(\boldn\) we solve by inspection the linear equation

The vectors \(\boldw_1=\angvec{2,-1,0}\) and \(\boldw_2=\angvec{3,0,1}\) are two such solutions, and clearly \(\boldw_1\ne c\boldw_2\) for any \(c\in \R\text{.}\) We conclude that

is a parametrization of \(\mathcal{S}\text{.}\)

Following Example 2.5.7, we easily see that \(\mathcal{S}_1\) is parametrized by \(\boldr(u,v)=(u,\sqrt{u^2+v^2})\)

If we wanted to settle for a piecewise parametrization of \(\mathcal{S}_2\) we could proceed exactly as above, parametrizing the right half of \(\mathcal{S}_2\) as \(\boldr_1(u,v)=(u,\sqrt{u^2+v^2})\) and the left half as \(\boldr_2(u,v)=(u,v,-(u^2+v^2))\text{.}\) Instead, we use a variant of spherical coordinates. For \(P=(x,y,z)\in \R^3\text{,}\) let \(\rho=\abs{(x,y,z)}\text{,}\) let \(\phi\) be the angle between \(\overrightarrow{OP}\) and the positive \(y\)-axis (as opposed to the \(z\)-axis), and let \(\theta\) be the angle between the positive \(z\)-axis and the projection of \(\overrightarrow{OP}\) onto the \(xz\)-plane. With this setup we have

(Essentially this is just the usual spherical coordinates along with the “cyclical” renaming of the axes \(y\rightarrow z, z\rightarrow x, x\rightarrow y\text{.}\)) With respect to these coordinates, the half-cone \(\mathcal{S}_1\) can be described implicitly as \(\phi=\pi/4\text{,}\) \(0\leq \rho\text{.}\) To get the full cone \(\mathcal{S}_2\) we simply remove the restriction that \(\rho\) be nonnegative. This gives rise to the parametrization

We compute

At \(P=(1,1,1)=\boldr(1,1)\) we have

This \(\angvec{-2,-2,4}\) is a normal vector of the tangent plane. We can always scale this by any nonzero constant to get another, more convenient normal vector: e.g., \(\boldn=-\frac{1}{2}\angvec{-2,-2,4}=\angvec{1,1,-2}\text{.}\) Lastly, the plane passing through \(P=(1,1,1)\) with normal vector \(\boldn=\angvec{1,1,-2}\) has defining equation

1. First we parametrize \(\mathcal{S}\) using

   \begin{align*} \boldr(\phi,\theta)\amp =(R\sin\phi\cos\theta, R\sin\phi\sin\theta, R\cos\phi)\\ \mathcal{R}\amp\colon 0\leq \phi\leq \pi, 0\leq \theta\leq 2\pi\text{,} \end{align*}

   as in Example 2.5.5.

2. Now assemble ingredients:

   \begin{align*} \boldr_\phi \amp=\langle R\cos\phi\cos\theta, R\cos\phi\sin\theta, -R\sin\phi\rangle \\ \boldr_\theta\amp=\langle-R\sin\phi\sin\theta, R\sin\phi\sin\theta, 0\rangle\\ \boldr_\phi\times \boldr_\theta\amp=R^2\sin\phi\angvec{\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi} \\ \abs{\boldr_\phi\times\boldr_\theta}\amp =\val{R^2\sin\phi}\abs{\angvec{\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi}}\\ \amp=R^2\sin\phi\text{,} \end{align*}

   where the last equality follows since \(\sin\phi\geq 0\) for \(\phi\in [0,\pi]\) and \(\angvec{\sin\phi\cos\theta, \sin\phi\sin\theta,\cos\phi}\) parametrizes points on the unit sphere!

3. Finally, we integrate

   \begin{align*} \operatorname{area} \mathcal{S} \amp=\iint\limits_\mathcal{R}R^2\sin\phi\, dA \\ \amp=\int_0^\pi\int_0^{2\pi}R^2\sin\phi\, d\theta\, d\phi \\ \amp = 2\pi\, R^2(-\cos\phi)\Bigr\vert_0^{\pi}\\ \amp = 4\pi R^2\text{.} \end{align*}

1. First parametrize, using cylindrical coordinates:

   \begin{align*} \bolds(r, \theta) \amp = (r\cos\theta, r\sin\theta, r^2)\\ \mathcal{R}\amp\colon 0\leq r\leq 2, 0\leq \theta\leq 2\pi \text{.} \end{align*}

   Here the region \(\mathcal{R}\) is a polar description of the projection of \(\mathcal{S}\) on the \(xy\)-plane, obtained by first observing that the intersection of the paraboloid with the plane \(z=4\) is given by \(z=r^2=4\text{,}\) or \(r=2\text{.}\)

2. Now assemble ingredients:

   \begin{align*} \bolds_r \amp = \langle\cos\theta, \sin\theta, 2r\rangle)\\ \bolds_\theta \amp= \angvec{-r\sin\theta, r\cos\theta, 0}\\ \bolds_r\times \bolds_\theta \amp=\angvec{-2r^2\cos\theta, -2r^2\sin\theta, r} \\ \abs{\bolds_r\times \bolds_\theta} \amp=\sqrt{4r^4\cos^2\theta+4r^4\sin^2\theta+r^2}\\ \amp = \sqrt{4r^4+r^2}=r\sqrt{4r^2+1}\text{.} \end{align*}

3. Finally, we integrate

   \begin{align*} \operatorname{area}\mathcal{S} \amp= \iint\limits_\mathcal{R}r\sqrt{4r^2+1}\, dA \\ \amp=\int_0^{2\pi}\int_0^2r\sqrt{4r^2+1}\, dr\, d\theta \\ \amp =2\pi \left(\frac{1}{12}(4r^2+1)^{3/2}\right)\Bigr\vert_0^2\\ \amp= \frac{\pi}{6}(17^{3/2}-1) \text{.} \end{align*}