- 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 7 of Unit 3.
Brain Gains (Rapid Recall, Jivin' Generation)
- Consider the solid region $D$ in the first octant that lies below the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{5}=1$.
- Set up an iterated triple integral to find the volume of $D$.
- Set up an iterated integral formula to give the $y$-coordinate of the centroid of $D$.
- The solid $D$ has a density given by $\delta = 2xz$ grams per cubic centimeter. Set up an iterated integral formula that would give $\bar z$, the $z$-coordinate of the center-of-mass of the pyramid.
Solution
The volume is given by $$\ds \int_{0}^{2}\int_{0}^{3(1-x/2)}\int_{0}^{5(1-x/2-y/3)}1 dzdydx.$$ The $y$=-coordinate of the centroid is $$\ds \bar y = \frac{\int_{0}^{2}\int_{0}^{3(1-x/2)}\int_{0}^{5(1-x/2-y/3)} y dzdydx}{\int_{0}^{2}\int_{0}^{3(1-x/2)}\int_{0}^{5(1-x/2-y/3)}1 dzdydx}.$$ The $z$=-coordinate of the center-of-mass is $$\ds \bar z = \frac{\int_{0}^{2}\int_{0}^{3(1-x/2)}\int_{0}^{5(1-x/2-y/3)} z (2xz) dzdydx}{\int_{0}^{2}\int_{0}^{3(1-x/2)}\int_{0}^{5(1-x/2-y/3)}(2xz) dzdydx}.$$ We can quickly compute all of these with Mathematica.
outerB = {x, 0, 2};
middleB = {y, 0, 3 (1 - x/2)};
innerB = {z, 0, 5 (1 - x/2 - y/3)};
R = ParametricRegion[{x, y, z}, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
volume = Integrate[1, outerB, middleB, innerB]
centroidy = Integrate[y, outerB, middleB, innerB]/Integrate[1, outerB, middleB, innerB]
density = 2 x z
centerofmassz = Integrate[y*density, outerB, middleB, innerB]/Integrate[density, outerB, middleB, innerB]
- A wire is coiled and lies along the path given by $\vec r(t) = (3\cos t, 3\sin t, 4t)$ for $0\leq t\leq 6\pi$. In unit 1 we found the length of the wire was computed using $$ \ds s = \int_Cds = \int_C\left|\frac{d\vec r}{dt}\right|dt = \int_{0}^{6\pi}\sqrt{(-3\sin t)^2+(3\cos t)^2+4^2}dt = \int_{0}^{6\pi} 5 dt. $$ The density at points along the wire is given by $\delta = z$ grams per meter (the wire gets thicker as you move up the helix). Set up an integral to compute the mass of the wire, and then setup an integral to find the $x$-coordinate of the center-of-mass of the wire.
Solution
The mass is given by $$\ds m = \int_C dm = \int_C \delta ds = \int_{0}^{6\pi} (4t) 5 dt. $$ The $x$-coordinate of the center-of-mass is $$\ds \bar x = \frac{ \int_C x dm }{ \int_C dm } =frac{ \int_{0}^{6\pi}(3\cos t) (4t) 5 dt}{ \int_{0}^{6\pi} (4t) 5 dt}. $$ We can compute all these in Mathematica.
tBounds = {t, 0, 6 Pi}
ParametricPlot3D[{3 Cos[t], 3 Sin[t], 4 t}, Evaluate[tBounds]]
Integrate[5, tBounds]
Integrate[(4 t) *5, tBounds]
Integrate[(3 Cos[t]) (4 t) *5, tBounds]/Integrate[(4 t) *5, tBounds]
- Let $\vec u = (1, 2, 3)$, $\vec v = (3, 0, -1)$ and $\vec w = (2, -1, 2)$. Compute the triple product $(\vec u\times\vec v)\cdot \vec w$. The absolute value of this quantity gives the volume of the parellelepiped formed by these 3 vectors.
Solution
The solution is -26. The volume of the parallelepiped formed by these three vectors is 26. We'll use this fact to perform 3D change-of-coordinates.
u = {1, 2, 3};
v = {3, 0, -1};
w = {2, -1, 2};
Cross[u, v]
Dot[Cross[u, v], w]
Graphics3D[Parallelepiped[{0, 0, 0}, {u, v, w}], Axes -> True, AxesOrigin -> {0, 0, 0}, AxesLabel -> Automatic]
- Set up a triple integral that would give the volume of the solid region in space that is bounded by the surfaces $z = 2-y^2$, $y+z=0$, $x=0$ and $x=3$. Construct a rough sketch of the region as part of your work.
Solution
We need to find where the curves $y+z=0$ and $z=2-y^2$ intersect. Graphing them by hand and plotting points carefully is one option to see they intersection at $(y,z)=(-1,2)$ and $(y,z)=(2,-2)$. Alternately, we can solve the system of equations (solve each for $z$ which then yields $-y=2-y^2$, which is a quadratic we can solve to get $y=-1$ or $y=2$).
An integral for the volume of the solid region is $$\ds\int_{0}^{3}\int_{-1}^{2}\int_{-y}^{2-y^2}dzdydx.$$
Here's the region, with Mathematica.
outerB = {x, 0, 3};
middleB = {y, -1, 2};
innerB = {z, -y, 2 - y^2};
R = ParametricRegion[{x, y, z}, Evaluate[{outerB, middleB, innerB}]];
Region[R, Axes -> True, AxesLabel -> {x, y, z},
AxesOrigin -> {0, 0, 0}]
Group Problems
- Compute the volume of the parallelepiped whose edges are given by the vectors $\vec u = (1, 0, 1)$, $\vec v = (0, -1, 2)$ and $\vec w = (-2, 1, 0)$. [Check: 4]
- Set up an iterated triple integral to find the volume inside the sphere $x^2+y^2+z^2=9$. You can check that you've got the right bounds with Mathematica, or just compare with your peers.
- Draw the 3D solid that lies above the surface $z=\sqrt{x^2+y^2}$ and below the plane $z=3$. Then set up a triple integral formula to compute the $z$-coordinate of the centroid of the object.
- A wire is parametrized by $\vec r(t) = (2\cos t, t^2, 3t)$ for $0\leq t\leq 7$. The temperature at points on the wire is given by $T(x,y,z)=x^2+yz$. Set up an iterated integral formula that would give the average temperature of the wire.
- Set up a triple integral that would give the volume of the solid region that is bounded above by the paraboloid $z=6-x^2-y^2$, and below by the cone $z = \sqrt{x^2+y^2}$. Construct a rough sketch of the region as part of your work.
- Set up a triple integral that would give the volume of the solid region in the first octant that is below the plane $z=9$ and above the paraboloid $z=x^2+y^2$.
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