This is day 9 of Unit 3.

Brain Gains (Rapid Recall, Jivin' Generation)

  • Give cylindrical equations for both the the sphere $x^2+y^2+z^2=9$ and paraboloid $z=9-x^2-y^2$.

Solution

Recall that $x^2+y^2=r^2$.

  • An equation for the sphere is $r^2+z^2=9$.
  • An equation for the paraboloid is $z=9-r^2$.
  • Set up an iterated integral formula in spherical coordinates ($x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$) to find the $x$-coordinate of the centroid of the solid domain $D$ in the first octant that lies above the cone $x^2+y^2=z^2$ and below the sphere $x^2+y^2+z^2=49$.

Solution

The volume is given by $\ds V= \int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta$, which means $$\bar x = \frac{\iiint_D x dV}{\iiint_D dV}= \frac{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}(\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta}.$$ The following Mathematica code will visualize the region, and compute the centroid. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[rho Sin[phi] Cos[theta] rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]/
 Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]
  • Draw the region whose volume is given by $\ds \int_{0}^{\pi}\int_{1}^{4}\int_{1}^{z}r dr dz d\theta$.

Solution 

outerB = {theta, 0, Pi}
middleB = {z, 1, 4}
innerB = {r, 1, z}

coordinates = {r Cos[theta], r Sin[theta], z}
R = ParametricRegion[coordinates, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
  • Draw the region whose volume is given by $\ds \int_{\pi/2}^{\pi}\int_{0}^{\pi/2}\int_{4}^{5}\rho^2\sin\phi d\rho d\phi d\theta$.

Solution 

outerB = {theta, Pi/2, Pi}
middleB = {phi, 0, Pi/2}
innerB = {rho, 4, 5}

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
  • Consider the region $R$ in the first quadrant that lies above the curves $y=x$ and $y=\frac{1}{x}$, and below the curves $y=3x$ and $y=\frac{4}{x}$. Note that we can rewrite $x\leq y\leq 3x$ as $1\leq \frac{y}{x}\leq 3$, and we can rewrite $\frac{1}{x}\leq y\leq \frac{4}{x}$ as $1\leq xy\leq 4$. Using the change-of-coordinates $u = xy$ and $v=\frac{y}{x}$, compute the integral $\ds\iint_R dA$.

Solution

Computing differentials gives us $$\begin{pmatrix}du\\dv\end{pmatrix} = \begin{pmatrix}y\\-y/x^2\end{pmatrix}dx+\begin{pmatrix}x\\1/x\end{pmatrix}dy = \begin{bmatrix}y&x\\-y/x^2&1/x\end{bmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}.$$ The Jacobian of the transformation is $$\frac{\partial (u,v)}{\partial(x,y)} = |(y)(1/x) - (-y/x^2)(x)| = |2(y/x)| = |2v| = 2v.$$ We need $\ds \frac{\partial (x,y)}{\partial(u,v)} = \frac{1}{2v},$ which gives $$\iint_R dA = \iint \frac{\partial (x,y)}{\partial(u,v)} dudv = \int_{1}^{3}\int_{1}^{4}\frac{1}{2v}dudv.$$ Software can quickly compute this integral. Here is some Mathematica code that draws the region, and then computes the integral in three different ways. 

R = ImplicitRegion[1 <= x y <= 4 && 1 <= y/x <= 3 && x >= 0, {x, y}]
Show[Plot[{1/x, 4/x, 3 x, x}, {x, 0, 4}, PlotRange -> {0, 4}], Region[R]]

(*Integrate using the change-of-coordiantes.*)
Integrate[1/(2 v), {v, 1, 3}, {u, 1, 4}] // N

(*Integrate using Mathematica's region utilities.*)
Integrate[1, {x, y} \[Element] R] // N

(*Compute the integral by splitting the region up into 3 Cartesian regions.*)
Integrate[1, {x, Sqrt[1/3], 1}, {y, 1/x, 3 x}] + 
 Integrate[1, {x, 1, Sqrt[4/3]}, {y, x, 3 x}] + 
 Integrate[1, {x, Sqrt[4/3], 2}, {y, x, 4/x}] // N

Here is a way to visualize the region that shows the images of 1 by 1 boxes from the $uv$-plane, and 1/4 by 1/4 boxes, so we can visualize what the Jacobian $\frac{1}{2v}$ actually means. 

coordinates = {Sqrt[u /v], Sqrt[u v]};
ParametricPlot[Evaluate[coordinates, {u, 1, 4}, {v, 1, 3}], Mesh -> 1 {3, 2} - {1, 1}]

coordinates = {Sqrt[u /v], Sqrt[u v]};
ParametricPlot[Evaluate[coordinates, {u, 1, 4}, {v, 1, 3}], Mesh -> 4 {3, 2} - {1, 1}]

Group Problems

  1. The spherical change-of-coordinates is given by $$(x,y,z) = (\underbrace{\rho\sin\phi}_{r}\cos\theta, \underbrace{\rho\sin\phi}_{r}\sin\theta, \rho\cos\phi).$$
    1. Give Cartesian coordinates $(x,y,z)$ for the spherical coordinates $(\rho,\phi,\theta)$ given by $(2,\pi/2,\pi)$, $(2,\pi,\pi/2)$, and $(2,0,3\pi)$.
    2. Explain why an equation of the sphere $x^2+y^2+z^2=9$ in spherical coordinates is $\rho = 3$.
    3. Explain why an equation of the cone $x^2+y^2=z^2$ (so $r^2=z^2$ or $r=z$) in spherical coordinates is $\phi = \pi/4$.
    4. Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the sphere $x^2+y^2+z^2=9$ and below by the cone $z^2=x^2+y^2$. The Jacobian for spherical coordinates is $|\rho^2\sin\phi|$.
      • You can check your solution with Mathematica by updating the code below. 
        coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{rho, 0, 3}, {phi, Pi/4, Pi/2}, {theta, 0, Pi}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
    5. Explain why an equation of the plane $z=8$ in spherical coordinates is $\rho = 8\sec \phi$.
    6. Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the plane $z=8$ and below by the cone $z^2=x^2+y^2$. Check your work with Mathematica.

Some Solutions

Here is the region for part d. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta.$$ 

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}]

Here is the region for part f. 

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8/ \cos\phi}\rho^2\sin\phi d\rho d\phi d\theta.$$ 

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}]


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