Math 214 - Multivariable Calculus - Class

Brain Gains (Rapid Recall, Jivin' Generation)

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}]

Consider the region $R$ that lies inside the triangle with vertices $(0,1)$, $(1,-2)$, and $(2,3)$, and the change-of-coordinates $u=x-y+1$, $v=3x+y-1$. (These coordinates were obtained by finding an equation of the lines of two edges of the triangle.) Show that $\ds \frac{\partial (x,y)}{\partial(u,v)} = \frac{1}{4}$, and then fill in the missing bounds of the integral $$\ds\iint_R 3x+y dA = \int_{0}^{?}\int_{0}^{?}(v+1)(\frac{1}{4})dvdu.$$

Solution

The Jacobian of $u$ and $v$ with respect to $x$ and $y$ is the area of the parallelogram with edges equal to $\frac{\partial (u,v)}{\partial x}=(1,3)$ and $\frac{\partial (u,v)}{\partial y}=(-1,1)$, which is $A = |(1)(1)-(3)(-1)|=4$. This gives $\ds \frac{\partial (u,v)}{\partial(x,y)} = 4$, and hence we have $\ds \frac{\partial (x,y)}{\partial(u,v)} = \frac{1}{4}$.

To find the bounds, we need to figure out where the points $(0,1)$, $(1,-2)$, and $(2,3)$ lie in the $(u,v)$ plane.

The point $(x,y)=(0,1)$ is at $u=0-1+1 = 0$ and $v=3(0)+1-1 = 0$, so $(u,v)=(0,0)$, the origin of our new coordinate system.
The point $(x,y)=(1,-2)$ is at $u=1-(-2)+1 = 4$ and $v=3(1)+(-2)-1 = 0$, so $(u,v)=(4,0)$.
The point $(x,y)=(2,3)$ is at $u=2-(3)+1 = 4$ and $v=3(2)+(3)-1 = 0$, so $(u,v)=(0,8)$.

The three points form a triangle in the $uv$-plane. The top edge of that triangle is the line $v=8-2u$. As such, we have $$\ds\iint_R 3x+y dA = \int_{0}^{4}\int_{0}^{8-2u}(v+1)(\frac{1}{4})dvdu.$$

coordinates = {(u + v)/4, (v - 3 u + 4)/4};
ParametricPlot[Evaluate[coordinates, {u, 0, 4}, {v, 0, 8 - 2 u}], Mesh -> {10, 0}]

A bead is formed by drilling a hole through a sphere. The solid region shown below is the region inside a sphere of radius 2 and outside a cylinder of radius 1. Set up an iterated triple integral in spherical coordinates that would give the volume of the solid.

Solution

The sphere has equation $\rho = 2$. The cylinder has equation $r=1$ or rather $\rho \sin \phi = 1$, which we can rewrite at $\rho = \csc \phi$. The two surfaces intersect when $\sin\phi = \frac{1}{2}$, so when $\phi = \pi/6$ and $\phi = 5\pi/6$. The volume is given by the integral $$V=\int_{0}^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc \phi}^{2}\rho^2\sin\phi d\rho d\phi d\theta.$$

We can check our solution is correct by drawing the region with Mathematica. The code below uses the custom function plotRegion3D[]. You can also use the built in ParametricRegion[] command (see a previous day for examples).

plotRegion3D[cs_, ob_, mb_, ib_] := 
 Show[{{ParametricPlot3D[Evaluate[Table[cs /. (ib[[1]] -> ib[[i]]), {i, 2, 3}], ob, mb], AxesLabel -> {x, y, z}, Mesh -> {15, 1}], 
    ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> mb[[i]]), {i, 2, 3}], ob], {s, 0, 1}, PlotStyle -> {Red, Blue}, Mesh -> {15, 0}], 
    ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> (mb[[2]] (1 - t) + mb[[3]] t)) /. (ob[[1]] -> ob[[i]]), {i, 2, 3}]], {t, 0, 1}, {s, 0, 1}, PlotStyle -> Green, Mesh -> {0, 0}]}}, PlotRange -> All]

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
plotRegion3D[coordinates, {theta, 0, 2 Pi}, {phi, Pi/6, 5 Pi/6}, {rho, Csc[phi], 2}]

Consider the 2D region in the $yz$-plane satisfying $1\leq y \leq 2$ and $y\leq z\leq 9-y^2$. This region is revolved about the $z$-axis to produce a solid of revolution. Set up a triple integral in cylindrical coordinates that would give the volume of this solid. Half of the region is shown below.

Solution

The volume is $$V=\int_{0}^{2\pi}\int_{1}^{2}\int_{r}^{9-r^2}r dz dr d\theta.$$

We can check our solution is correct by drawing the region with Mathematica, using the custom function plotRegion3D[] (defined in the previous question).

coordinates = {r Cos[theta], r Sin[theta], z}
plotRegion3D[coordinates, {theta, 0, 2 Pi}, {r, 1, 2}, {z, r, 9-r^2}]

Group Problems

Consider the solid of revolution obtained by revolving about the $z$-axis the 2D region in the $xz$-plane that lies below the line $z = 4-x$ for $2\leq x\leq 4$.
1. Set up an iterated triple integral in cylindrical coordinates using the order $d\theta dz dr$ that would give the volume of this solid.
2. Set up an iterated triple integral in cylindrical coordinates using the order $d\theta dr dz$ that would give the volume of this solid.
3. Use Mathematica to verify that the bounds you gave do indeed correctly describe the solid.
4. Compute one of the integrals above by hand (feel free to check with software).
Consider the region $R$ in the plane that lies inside the triangle with corners $(0,0)$, $(1,2)$, and $(-3,1)$. Two edges are given by the lines $2x-y=0$ and $x+3y=0$, so let's use the change-of-coordinates $u=2x-y$ and $v=x+3y$.
1. Compute $\ds \frac{\partial (u,v)}{\partial(x,y)}$ and $\ds \frac{\partial (x,y)}{\partial(u,v)}$.
2. Draw the region $R$ in the $uv$-plane.
3. Use the change-of-coordinates to set up an appropriate integral in terms of $u$ and $v$ to compute $\iint_R xy dA$.
Consider the vector field $\ds \vec F = \frac{1}{2\pi}\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.
1. Let $C$ be the curve parametrized by $\vec r(t) = (3\cos t, 3\sin t)$ for $0\leq t\leq 2\pi$. Compute the work done by $\vec F$ to move an object along $C$. [Hint: The answer is NOT zero.]
2. Let $C$ be the curve parametrized by $\vec r(t) = (7\cos t, 7\sin t)$ for $0\leq t\leq 4\pi$. Compute the work done by $\vec F$ to move an object along $C$.
3. Let $C$ be the curve parametrized by $\vec r(t) = (a\cos t, a\sin t)$ for $0\leq t\leq 2k\pi$. Compute the work done by $\vec F$ to move an object along $C$.
4. Show that $D\vec F(x,y)$ is symmetric.
5. Does $\vec F$ have a potential?

Solutions

If $\vec F$ had a potential, then the work done along a closed curve would be zero. This is not true, as the first 3 parts illustrate. The vector field does NOT have a potential.

The function and derivative are not defined at $(0,0)$, which means the domain of the vector field is not simply connected. We cannot use the symmetry of the derivative to conclude that there is a potential, because the domain is not simply connected.

Here are some Mathematica computations relevant to this problem.

ClearAll[F, r]
F[x_, y_] := 1/(2 Pi (x^2 + y^2)) {-y, x}
r[t_] := {a Cos[t], a Sin[t]}

(*Is the derivative symmetric*)
D[F[x, y], {{x, y}}] // Simplify // MatrixForm

(*Try to find a potential*)
Integrate[F[x, y][[1]], x]
Integrate[F[x, y][[2]], y]

(*Compute work done by wrapping k times around a circle of radius a*)
Dot[F @@ r@t, D[r[t], t]] // Simplify
Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 k Pi}]

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HomePage Schedule	Class Day 39 Brain Gains (Rapid Recall, Jivin' Generation) 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}] Consider the region $R$ that lies inside the triangle with vertices $(0,1)$, $(1,-2)$, and $(2,3)$, and the change-of-coordinates $u=x-y+1$, $v=3x+y-1$. (These coordinates were obtained by finding an equation of the lines of two edges of the triangle.) Show that $\ds \frac{\partial (x,y)}{\partial(u,v)} = \frac{1}{4}$, and then fill in the missing bounds of the integral $$\ds\iint_R 3x+y dA = \int_{0}^{?}\int_{0}^{?}(v+1)(\frac{1}{4})dvdu.$$ Solution The Jacobian of $u$ and $v$ with respect to $x$ and $y$ is the area of the parallelogram with edges equal to $\frac{\partial (u,v)}{\partial x}=(1,3)$ and $\frac{\partial (u,v)}{\partial y}=(-1,1)$, which is $A = \|(1)(1)-(3)(-1)\|=4$. This gives $\ds \frac{\partial (u,v)}{\partial(x,y)} = 4$, and hence we have $\ds \frac{\partial (x,y)}{\partial(u,v)} = \frac{1}{4}$. To find the bounds, we need to figure out where the points $(0,1)$, $(1,-2)$, and $(2,3)$ lie in the $(u,v)$ plane. The point $(x,y)=(0,1)$ is at $u=0-1+1 = 0$ and $v=3(0)+1-1 = 0$, so $(u,v)=(0,0)$, the origin of our new coordinate system. The point $(x,y)=(1,-2)$ is at $u=1-(-2)+1 = 4$ and $v=3(1)+(-2)-1 = 0$, so $(u,v)=(4,0)$. The point $(x,y)=(2,3)$ is at $u=2-(3)+1 = 4$ and $v=3(2)+(3)-1 = 0$, so $(u,v)=(0,8)$. The three points form a triangle in the $uv$-plane. The top edge of that triangle is the line $v=8-2u$. As such, we have $$\ds\iint_R 3x+y dA = \int_{0}^{4}\int_{0}^{8-2u}(v+1)(\frac{1}{4})dvdu.$$ coordinates = {(u + v)/4, (v - 3 u + 4)/4}; ParametricPlot[Evaluate[coordinates, {u, 0, 4}, {v, 0, 8 - 2 u}], Mesh -> {10, 0}] A bead is formed by drilling a hole through a sphere. The solid region shown below is the region inside a sphere of radius 2 and outside a cylinder of radius 1. Set up an iterated triple integral in spherical coordinates that would give the volume of the solid. Solution The sphere has equation $\rho = 2$. The cylinder has equation $r=1$ or rather $\rho \sin \phi = 1$, which we can rewrite at $\rho = \csc \phi$. The two surfaces intersect when $\sin\phi = \frac{1}{2}$, so when $\phi = \pi/6$ and $\phi = 5\pi/6$. The volume is given by the integral $$V=\int_{0}^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc \phi}^{2}\rho^2\sin\phi d\rho d\phi d\theta.$$ We can check our solution is correct by drawing the region with Mathematica. The code below uses the custom function plotRegion3D[]. You can also use the built in ParametricRegion[] command (see a previous day for examples). plotRegion3D[cs_, ob_, mb_, ib_] := Show[{{ParametricPlot3D[Evaluate[Table[cs /. (ib[[1]] -> ib[[i]]), {i, 2, 3}], ob, mb], AxesLabel -> {x, y, z}, Mesh -> {15, 1}], ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> mb[[i]]), {i, 2, 3}], ob], {s, 0, 1}, PlotStyle -> {Red, Blue}, Mesh -> {15, 0}], ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> (mb[[2]] (1 - t) + mb[[3]] t)) /. (ob[[1]] -> ob[[i]]), {i, 2, 3}]], {t, 0, 1}, {s, 0, 1}, PlotStyle -> Green, Mesh -> {0, 0}]}}, PlotRange -> All] coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]} plotRegion3D[coordinates, {theta, 0, 2 Pi}, {phi, Pi/6, 5 Pi/6}, {rho, Csc[phi], 2}] Consider the 2D region in the $yz$-plane satisfying $1\leq y \leq 2$ and $y\leq z\leq 9-y^2$. This region is revolved about the $z$-axis to produce a solid of revolution. Set up a triple integral in cylindrical coordinates that would give the volume of this solid. Half of the region is shown below. Solution The volume is $$V=\int_{0}^{2\pi}\int_{1}^{2}\int_{r}^{9-r^2}r dz dr d\theta.$$ We can check our solution is correct by drawing the region with Mathematica, using the custom function plotRegion3D[] (defined in the previous question). coordinates = {r Cos[theta], r Sin[theta], z} plotRegion3D[coordinates, {theta, 0, 2 Pi}, {r, 1, 2}, {z, r, 9-r^2}] Group Problems Consider the solid of revolution obtained by revolving about the $z$-axis the 2D region in the $xz$-plane that lies below the line $z = 4-x$ for $2\leq x\leq 4$. Set up an iterated triple integral in cylindrical coordinates using the order $d\theta dz dr$ that would give the volume of this solid. Set up an iterated triple integral in cylindrical coordinates using the order $d\theta dr dz$ that would give the volume of this solid. Use Mathematica to verify that the bounds you gave do indeed correctly describe the solid. Compute one of the integrals above by hand (feel free to check with software). Consider the region $R$ in the plane that lies inside the triangle with corners $(0,0)$, $(1,2)$, and $(-3,1)$. Two edges are given by the lines $2x-y=0$ and $x+3y=0$, so let's use the change-of-coordinates $u=2x-y$ and $v=x+3y$. Compute $\ds \frac{\partial (u,v)}{\partial(x,y)}$ and $\ds \frac{\partial (x,y)}{\partial(u,v)}$. Draw the region $R$ in the $uv$-plane. Use the change-of-coordinates to set up an appropriate integral in terms of $u$ and $v$ to compute $\iint_R xy dA$. Consider the vector field $\ds \vec F = \frac{1}{2\pi}\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$. Let $C$ be the curve parametrized by $\vec r(t) = (3\cos t, 3\sin t)$ for $0\leq t\leq 2\pi$. Compute the work done by $\vec F$ to move an object along $C$. [Hint: The answer is NOT zero.] Let $C$ be the curve parametrized by $\vec r(t) = (7\cos t, 7\sin t)$ for $0\leq t\leq 4\pi$. Compute the work done by $\vec F$ to move an object along $C$. Let $C$ be the curve parametrized by $\vec r(t) = (a\cos t, a\sin t)$ for $0\leq t\leq 2k\pi$. Compute the work done by $\vec F$ to move an object along $C$. Show that $D\vec F(x,y)$ is symmetric. Does $\vec F$ have a potential? Solutions If $\vec F$ had a potential, then the work done along a closed curve would be zero. This is not true, as the first 3 parts illustrate. The vector field does NOT have a potential. The function and derivative are not defined at $(0,0)$, which means the domain of the vector field is not simply connected. We cannot use the symmetry of the derivative to conclude that there is a potential, because the domain is not simply connected. Here are some Mathematica computations relevant to this problem. ClearAll[F, r] F[x_, y_] := 1/(2 Pi (x^2 + y^2)) {-y, x} r[t_] := {a Cos[t], a Sin[t]} (Is the derivative symmetric) D[F[x, y], {{x, y}}] // Simplify // MatrixForm (Try to find a potential) Integrate[F[x, y][[1]], x] Integrate[F[x, y][[2]], y] (Compute work done by wrapping k times around a circle of radius a) Dot[F @@ r@t, D[r[t], t]] // Simplify Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 k Pi}] Edit Page History Source Attach File Backlinks List Group Page last modified on November 17, 2022, at 04:24 PM
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Day 39

Brain Gains (Rapid Recall, Jivin' Generation)

Group Problems