- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is day 3 of Unit 3.
Brain Gains (Rapid Recall, Jivin' Generation)
- What is the difference between the three double integrals $\iint_RdA$, $\iint_Rdydx$, and $\iint_Rdxdy$?
Solution
The three integral notations all add up area.
- The first notation $\iint_RdA$ assumes you have cut the region up into tiny rectangles where both the width and height have been reduced.
- The second notation $\iint_Rdydx$ slices a region up into vertical slabs whose height is $\int_C dy$ with a tiny width $dx$.
- The third notation $\iint_Rdxdy$ slices a region up into horizontal slabs whose width is $\int_C dx$ with a tiny height $dy$.
All three integrals aim to compute the same number, just in a different way. Fubini's theorem provides precise details under which these three integrals are guaranteed to be equal.
We can use the Mesh command in Mathematica to quickly visualize this.
outerB = {x, 0, 3};
innerB = {y, 0, 9 - x^2};
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {20, 20}]
ParametricPlot[coordinates, outerB, innerB, Mesh -> {20, 0}]
ParametricPlot[coordinates, outerB, innerB, Mesh -> {0, 20}]
- Consider the region in the first quadrant that lies left of the line $y=x/2$ and below the line $y=1$. A metal plate occupies this region, and has density given by $\delta = \cos(y^2)$.
- Set up an iterated integral of the form $\int_?^?\int_?^? ? dydx$ that would give the mass of the plate.
- Set up an iterated integral of the form $\int_?^?\int_?^? ? dxdy$ that would give the mass of the plate.
- Examine the two integrals above. One of them has a problem and cannot be computed. Which one, and why?
- Set up an integral formula to compute $\bar x$, the $x$-coordinate of the center-of-mass of the metal plate.
Solutions
Start by drawing the region, to discover that it's a triangle. Note that the intersection of $y=x/2$ (so $x=2y$) and $y=1$ is at $(2,1)$. The two integrals are $\ds\int_{0}^{2}\int_{x/2}^{1}\cos(y^2)dydx$ and $\ds\int_{0}^{1}\int_{0}^{2y}\cos(y^2)dydx$.
The first integral has a problem in that the inside integral cannot be done, as seen with the first block of code below. Mathematica has been coded in a way that it can recognize this issue, for this problem, and is still able to give an exact answer for the double integral. Putting an N in front of integrate swaps from getting an exact answer to a numerical approximation.
Integrate[Cos[y^2], {y, x/2, 1}]
Integrate[Cos[y^2], {x, 0, 2}, {y, x/2, 1}]
NIntegrate[Cos[y^2], {x, 0, 2}, {y, x/2, 1}]
The second integral can be done with substitution.
Integrate[Cos[y^2], {x, 0, 2 y}]
Integrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]
The center of mass formula, which you can give as either $\ds \bar x = \frac{\ds\int_{0}^{2}\int_{x/2}^{1}x\cos(y^2)dydx}{\ds\int_{0}^{2}\int_{x/2}^{1}\cos(y^2)dydx}$ or $\bar x = \frac{\ds\int_{0}^{1}\int_{0}^{2y}x\cos(y^2)dydx}{\ds\int_{0}^{1}\int_{0}^{2y}\cos(y^2)dydx}$, requires numerical integration. Let's look at Mathematica's solution.
Integrate[x Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]/Integrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]
NIntegrate[x Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]/NIntegrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]
We can utilize the same code we've been using the last few days to automate a lot of this, and draw the region (with appropriate mesh lines) at the same time.
outerB = {y, 0, 1}
innerB = {x, 0, 2 y}
density = Cos[y^2];
mass = Integrate[density, outerB, innerB]
xbar = Integrate[x*density, outerB, innerB]/mass
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
outerB = {x, 0, 2}
innerB = {y, x/2, 1}
density = Cos[y^2];
mass = Integrate[density, outerB, innerB]
xbar = Integrate[x*density, outerB, innerB]/mass
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
- Consider the change-of-coordinates $x = u-3v$ and $y = 2u+4v$.
- Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$.
- Compute the differential $(dx,dy)$ and write it as a linear combination $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}?\\?\end{pmatrix}du + \begin{pmatrix}?\\?\end{pmatrix}dv.$$
- Compute the area of the parallelogram you just drew using the area formula $|ad-bc|$. We call this the Jacobian of the change-of-coordinates, written $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
Solution
The region is a parallelogram with one vertex at the origin, and the two edges connecting to the origin end at $(1,2)$ and $(-3,4)$. WE can find these points by using the $(u,v)$ coordinates $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$.
For the differentials, we have $dx = 1du-3dv$ and $dy = 2du+4dv$. Arranging these as a linear combination gives $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}du + \begin{pmatrix}-3\\4\end{pmatrix}dv.$$ Note that the partial derivatives are precisely the same vectors which form the edges of the parallelogram.
The area of the parallelogram is $A = |(1)(4) - (2)(-3)| = 10$.
With Mathematica, we can draw the region, compute the derivative of the change-of-coordinates, and then use the determinant command to get the area of the parallelogram we drew.
coordinates = {u - 3 v, 2 u + 4 v}
R = ParametricRegion[coordinates, {{u, 0, 1}, {v, 0, 1}}];
Region[R, Axes -> True, AxesLabel -> {x, y}, AxesOrigin -> {0, 0}]
Dcoordinates = D[coordinates, {{u, v}}]
Dcoordinates // MatrixForm
Dcoordinates // Det // Abs
Group Problems
- Consider the region in the first quadrant that lies below the curve $y=\sqrt{x}$ for $0\leq x\leq 9$
- Set up iterated double integrals to find the area of the region using the order $dydx$.
- Set up iterated double integrals to find the area of the region using the order $dxdy$.
- Remember to check that you are correct with Mathematica (draw the region that your bounds give).
outerB = {x, ,} innerB = {y, ,} coordinates = {x, y}; ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}] outerB = {y, ,} innerB = {x, ,} coordinates = {x, y}; ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
- Consider the change-of-coordinates $x = -2u+3v$, $y=u+5v$.
- Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$.
- Compute the area of the region you just drew (the region should be a parallelogram).
- Compute the differential of the change-of-coordinates, and write it in the form $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
- State the Jacobian $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
- A box lies inside the rectangle $ [-2,6]\times [1,5] $ (so $-2\leq x\leq 6$ and $1\leq y \leq 5$ ).
- Use geometric reasoning to state the center-of-mass $(\bar x,\bar y)$ of the rectangle. (Where is the geometric center?)
- Compute the integral formula $\bar x = \ds\frac{\int_{-2}^{6}\int_1^5 x dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 64/32=2.]
- Compute the integral formula $\bar y = \ds\frac{\int_{-2}^{6}\int_1^5 y dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$ [Check: 96/32=3.]
- Find the area of the triangle whose vertices are $(-2,1)$, $(3,4)$, and $(1,7)$. [Check: 21/2]
- Pick values for $a,b,c,d$ in the change-of-coordinates $x = au+bv$, $y=cu+dv$.
- Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$. You can use the Mathematica code from above to check your answer.
- Compute the area of the region you just drew (the region should be a parallelogram).
- Compute the differential of the change-of-coordinates, and write it in the form $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
- State the Jacobian $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
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