logistic_guy
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Find the area of the base \(\displaystyle B\) of the solid.
for kids - 2
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Agent Smith
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[imath]15\pi = \frac{5}{3} \pi r^2[/imath]
logistic_guy
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\(\displaystyle \textcolor{indigo}{\bold{Correct.}}\) This is the volume.
The volume of the cone can be written as:
\(\displaystyle V = \frac{1}{3}Ah\)
Or
\(\displaystyle 15\pi = \frac{1}{3}A(5)\)
Then, the area of the base is:
\(\displaystyle A = 3\frac{1}{5}15\pi = \textcolor{blue}{9\pi \ \text{m}^2}\)
Agent Smith
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Thanks for posing easy questions too. Feeling really good about myself right now.
[imath]\pi \int_a ^b y^2 dx[/imath]
[imath]\pi \int_a ^b y^2 dx[/imath]
logistic_guy
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You're welcome.
It's either you're a smart kid or your mommy is sitting beside you guiding you.
This is an advanced formula. You will learn it once you reach college. Or sometimes even at high school.
How to use this integral precisely? You normally have the cone as in the picture of the OP. You flip it upside-down so that its base becomes in the top. Now the cone has its axis in \(\displaystyle z\) direction. You then rotate it \(\displaystyle 90^{\circ}\) to the right so that its axis lies in the positive \(\displaystyle x\)-axis.
The idea is to draw a slanted line and rotate it around the \(\displaystyle x\)-axis to get a full cone. If we try to locate two points on this slanted line, we get:
\(\displaystyle (0,0)\)
\(\displaystyle (h,r)\)
Then, you can construct a formula for this slanted line.
\(\displaystyle y - y_0 = m(x - x_0)\)
\(\displaystyle m = \frac{r - 0}{h - 0} = \frac{r}{h}\) is the slope of the slanted line.
Then the formula becomes:
\(\displaystyle y = \frac{r}{h}x\)
Now you can use your integral.
\(\displaystyle V = \pi\int_{a}^{b}y^2 \ dx = \pi\int_0^h \left(\frac{r}{h}x\right)^2 \ dx\)
\(\displaystyle = \frac{\pi r^2}{h^2}\frac{x^3}{3}\bigg|_{0}^{h} = \frac{\pi r^2}{h^2}\frac{h^3}{3} = \frac{\pi r^2h}{3}\)
Alternative ways to find the volume are:
\(\displaystyle V = \pi\int_a^b x^2 \ dy\)
\(\displaystyle V = 2\pi\int_a^b xy \ dx\)
\(\displaystyle V = 2\pi\int_a^b yx \ dy\)
Advanced methods to find the volume are:
\(\displaystyle V = \int dV = \int_{0}^{h}\int_{-\frac{r}{h}z}^{\frac{r}{h}z}\int_{-\sqrt{\left(\frac{r}{h}z\right) - x^2}}^{\sqrt{\left(\frac{r}{h}z\right) - x^2}} \ dy \ dx \ dz\)
\(\displaystyle V = \int dV = \int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{\frac{r}{h}(h - z)} r \ dr \ dz \ d\theta\)
Very advanced but also very fun method to find the volume of the cone is:
\(\displaystyle V = \int dV = \int_{0}^{2\pi}\int_{0}^{\tan^{-1}\frac{r}{h}}\int_{0}^{\frac{h}{\cos\phi}}\rho^2\sin\phi \ d\rho \ d\phi \ d\theta\)
Super advanced but also super fun method is:
\(\displaystyle V = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n} \ dS \quad \text{where} \quad \mathbf{F} = \frac{1}{3} \langle x, y, z \rangle\)
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Agent Smith
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Can all problems be reduced to
[imath]\frac{dy}{dx}[/imath]
[imath] \int_a ^b y dx[/imath]
[imath] \pi \int_a ^b y^2 dx[/imath]
[imath] \int_a ^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx[/imath]
[imath]2 \pi \int_a ^b y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx[/imath]
[imath]\frac{dy}{dx}[/imath]
[imath] \int_a ^b y dx[/imath]
[imath] \pi \int_a ^b y^2 dx[/imath]
[imath] \int_a ^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx[/imath]
[imath]2 \pi \int_a ^b y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx[/imath]
logistic_guy
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In your dreams. Have you wondered why there is \(\displaystyle \pi\) in your integrals?
Because they are solids of a revolution. In other words, you draw a line (not necessary straight) around an axis and you rotate it in a full circle to get the solid (volume).
For example, consider these two lines:
\(\displaystyle y = 1\)
\(\displaystyle y = x\)
If you rotate them around the positive \(\displaystyle x\)-axis for example, you get a cylinder and a cone respectively.
Let us analyze the line \(\displaystyle y = 1\) further. If you draw a vertical line from the \(\displaystyle x\)-axis to the line \(\displaystyle y = 1\) and rotate \(\displaystyle y = 1\) around the \(\displaystyle x\)-axis, the vertical line becomes a radius of a circle.
What is the length of this radius? It is always \(\displaystyle \textcolor{darkblue}{\bold{one}}\), no matter where you move on the \(\displaystyle x\)-axis. In simple words, the radius of the revolution is the function itself.
If we have a radius we can find the area of a circle, that is:
\(\displaystyle \pi y^2\), but wait if this area has a little height (thickness), it becomes a volume, then
\(\displaystyle \pi y^2 dx\)
This formula says that we have a cylinder with radius \(\displaystyle y\) and height \(\displaystyle dx\). That's where your first integral came from:
\(\displaystyle V = \pi \int_{a}^{b} y^2 \ dx\)
You can use the same formula to find the volume of a revolution of \(\displaystyle y = x\) which gives a volume of a cone with \(\displaystyle h = R\).
The more interesting topic is if we rotate \(\displaystyle y\) around the \(\displaystyle x\)-axis and we are interested only in the surface area of the solid.
\(\displaystyle 2\pi y\) is the circumference of one of the circles that creates the solid of revolution.
\(\displaystyle \text{circumference} \times \text{thickness} = \text{surface area}\)
\(\displaystyle \textcolor{blue}{2\pi y \ dx} \rightarrow\) gives a cross-section of the surface area of the cylinder.
Likewise
\(\displaystyle \textcolor{red}{2\pi y \ ds} \rightarrow\) gives a cross-section of the surface area of the cone.
We use the thickness of the cone as \(\displaystyle ds\) and not \(\displaystyle dx\) because \(\displaystyle y = x\) is not a horizontal line along the \(\displaystyle x\)-axis.
In fact:
\(\displaystyle ds^2 = dx^2 + dy^2\)
Or
\(\displaystyle ds = \sqrt{dx^2 + dy^2}\)
Then, we have a beautiful formula to compute the surface area of a solid of a revolution. That is:
\(\displaystyle S = 2\pi\int_{a}^{b} y \ ds = 2\pi\int_{a}^{b} y \ \sqrt{dx^2 + dy^2}\)
But wait, this notation is not the standard integral we know. Then, make it standard!
\(\displaystyle 1 = \frac{dx}{dx}\)
Then,
\(\displaystyle S = 2\pi\int_{a}^{b} y \ \sqrt{dx^2 + dy^2} \ \frac{dx}{dx} = 2\pi\int_{a}^{b} y \ \sqrt{\frac{dx^2 + dy^2}{dx^2}} \ dx\)
Or
\(\displaystyle S = 2\pi\int_{a}^{b} y \ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx\)
That's where your second integral came from.
This is a counter example where your integrals of revolution is not applicable.
If one day you were able to find the volume of this object, then you would know that you have mastered calculus perfectly!
logistic_guy
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And this tiny integral is just the area under the curve.
This integral becomes more interesting if it is written like this:
\(\displaystyle \int_a ^b y \ dA \rightarrow\) first moment of area about the \(\displaystyle x\)-axis.
Or
\(\displaystyle \int_a ^b y \ dm \rightarrow\) first moment of mass about the \(\displaystyle x\)-axis.
The first integral is related to the centroid of \(\displaystyle 2\text{D}\) objects such as the area of a triangle.
The second integral is related to the centroid of \(\displaystyle 3\text{D}\) objects such as a cone.
The coordinate of the centroid (center of mass) in the \(\displaystyle y\)-direction of a \(\displaystyle 3\text{D}\) object is:
\(\displaystyle \overline{y} = \frac{1}{M} \int y \ dm\)
Suppose we want to know how far is the centroid from the ground for the cone in the OP. In other words, we want to find the \(\displaystyle y\)-coordinate of the centroid.
To make the calculations easier, flip the cone in the OP upside-down so that its apex lies at the origin.
Take a cylindrical slice of the cone volume at height \(\displaystyle y\). That is:
\(\displaystyle dV = \pi r^2 \ dy\)
We also know that the mass density per unit volume is:
\(\displaystyle \rho = \frac{dm}{dV}\), then
\(\displaystyle dm = \rho \ dV\)
Then, our integral becomes:
\(\displaystyle \overline{y} = \frac{1}{M} \int y \ dm = \frac{1}{M} \int y \ \rho \ dV = \frac{1}{M} \int y \ \rho \ \pi r^2 \ dy \)
We can relate the distance \(\displaystyle y\) and the radius \(\displaystyle r\) by constructing similar right triangles.
\(\displaystyle \frac{r}{R} = \frac{y}{h}\)
Then,
\(\displaystyle r = \frac{R}{h}y\)
Our integral becomes:
\(\displaystyle \overline{y} = \frac{1}{M} \int y \ \rho \ \pi r^2 \ dy = \rho\frac{\pi R^2}{h^2M} \int_{0}^{h} y^3 \ dy\)
Or
\(\displaystyle \overline{y} = \rho\frac{\pi R^2}{h^2M} \frac{h^4}{4} = \rho\frac{\pi R^2}{M} \frac{h^2}{4}\)
But we already know that:
\(\displaystyle \rho = \frac{M}{V} = \frac{M}{\frac{1}{3}\pi R^2 h}\)
Then
\(\displaystyle \overline{y} = \rho\frac{\pi R^2}{M} \frac{h^2}{4} = \frac{M}{\frac{1}{3}\pi R^2 h}\frac{\pi R^2}{M} \frac{h^2}{4}\)
Or
\(\displaystyle \overline{y} = \frac{3}{4}h \rightarrow \) measured from the apex of the cone.
Flip the cone upside-down again. Then the \(\displaystyle y\)-coordinate of the center of mass of the cone in the OP is:
\(\displaystyle \overline{y} = \frac{1}{4}h \rightarrow\) measured from the base of the cone.
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logistic_guy
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@Agent Smith
Suppose that you got up in the morning bored and broke. You have no money and coincidently today is your wife's birthday and a month ago you promised yourself to buy her an expensive birthday present. Suddenly, you got a call from your Indian rich friend @khansaheb. You answered him and he has noticed that you are not okay from the tone of your voice. You explained to him your current situation and he promised to give you \(\displaystyle 2\)-million Indian rupees if you could solve the super integral to find out the volume of the cone in the OP. That is, you need to solve this:
\(\displaystyle V = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n} \ dS \quad \text{where} \quad \mathbf{F} = \frac{1}{3} \langle x, y, z \rangle\)
You have dug deep in your Calculus \(\displaystyle \text{III}\) notes, and you found out that the volume can be written as:
\(\displaystyle V = \iiint dV\)
Therefore, your first task was to prove:
\(\displaystyle \nabla \cdot \mathbf{F} = 1\)
Let \(\displaystyle \mathbf{F} = \langle F_1, F_2, F_3 \rangle = \frac{1}{3} \langle x, y, z \rangle\)
Then,
\(\displaystyle \nabla \cdot \mathbf{F} = \frac{1}{3}\left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\right) = \frac{1}{3}\bigg(1 + 1 + 1\bigg) = 1\)
Your first task has been done successfully. Now you can move to the second task which is to prove:
\(\displaystyle \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \frac{\pi r^2 h}{3}\)
You were very good in Calculus \(\displaystyle \text{III}\) and instantly you noticed that the cone can be described by this equation:
\(\displaystyle z = \frac{h}{R}\sqrt{x^2 + y^2}\)
Or
\(\displaystyle f(x,y) = z = \frac{h}{R}\sqrt{x^2 + y^2}\)
Then, the position vector is:
\(\displaystyle \bold{r} = \langle x, y, f(x,y) \rangle\)
And
\(\displaystyle \bold{n} = \frac{\bold{r}_x \times \bold{r}_y}{|\bold{r}_x \times \bold{r}_y|}\)
\(\displaystyle dS = |\bold{r}_x \times \bold{r}_y| \ dx \ dy\)
Let \(\displaystyle \bold{m}\) be the normal vector, then
\(\displaystyle \bold{m} = \bold{r}_x \times \bold{r}_y\)
Or
\(\displaystyle \bold{n} \ dS = \frac{\bold{m}}{|\bold{r}_x \times \bold{r}_y|}|\bold{r}_x \times \bold{r}_y| \ dx \ dy = \bold{m} \ dx \ dy\)
Then,
\(\displaystyle \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \iint_R \mathbf{F} \cdot \bold{m} \ dx \ dy\)
Your wife made you an orange juice and while you were drinking you noticed that:
\(\displaystyle z = \frac{h}{R}\sqrt{x^2 + y^2} = \frac{h}{R}r\)
You immediately called Khan and asked him if he may allow you to work in the Cylindrical coordinate as from your long experience you know that the Cartesian coordinate is not always friendly in solving integrals. Khan as a gentleman accepted.
Therefore, your integral becomes:
\(\displaystyle \iint_R \mathbf{F} \cdot \bold{m} \ dx \ dy = \iint_R \mathbf{F} \cdot \bold{m} \ dr \ d\theta\)
where the position vector is now:
\(\displaystyle \bold{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, z \rangle\)
So that \(\displaystyle \bold{m} = \bold{r}_{r} \times \bold{r}_{\theta}\)
You are very smart, so you noticed that the cone has two surface areas. One is the slant surface area while the other is the base of the cone.
Therefore, the integral Khan gave you becomes:
\(\displaystyle V = \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \iint_{R_1} \mathbf{F} \cdot \bold{m}_1 \ dr \ d\theta + \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
This means you have to work with two different position vectors:
\(\displaystyle \bold{r}_1(r,\theta) = \left\langle r\cos\theta, r\sin\theta, \frac{h}{R}r \right\rangle\)
\(\displaystyle \bold{r}_2(r,\theta) = \langle r\cos\theta, r\sin\theta, h \rangle\)
You decided to work with the first integral first:
\(\displaystyle V_1 = \iint_{R_1} \mathbf{F} \cdot \bold{m}_1 \ dr \ d\theta\)
Your task is now to find \(\displaystyle \bold{m}_1\).
\(\displaystyle \bold{m}_1 = \bold{r}_{1r} \times \bold{r}_{1\theta}\)
\(\displaystyle \bold{r}_{1r} = \frac{\partial \bold{r}_1}{\partial r} = \left\langle \cos\theta, \sin\theta, \frac{h}{R} \right\rangle\)
\(\displaystyle \bold{r}_{1\theta} = \frac{\partial \bold{r}_1}{\partial \theta} = \left\langle -r \sin\theta, r \cos\theta, 0 \right\rangle\)
Then,
\(\displaystyle \bold{m}_1 = \bold{r}_{1r} \times \bold{r}_{1\theta} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & \frac{h}{R} \\-r \sin\theta & r \cos\theta & 0\end{vmatrix} = \left\langle -\frac{h r}{R} \cos\theta, -\frac{h r}{R} \sin\theta, r \right\rangle\)
Now your task is to take the dot product. That is:
\(\displaystyle \bold{F} = \frac{1}{3} \langle x, y, z \rangle = \frac{1}{3} \langle r \cos\theta, r \sin\theta, \frac{h}{R} r \rangle\)
\(\displaystyle \bold{F} \cdot \bold{m}_1 = \frac{1}{3} \langle r \cos\theta, r \sin\theta, \frac{h}{R} r \rangle \cdot \left\langle -\frac{h r}{R} \cos\theta, -\frac{h r}{R} \sin\theta, r \right\rangle\)
\(\displaystyle = \frac{1}{3} \left( r \cos\theta \cdot \left(-\frac{h r}{R} \cos\theta\right) + r \sin\theta \cdot \left(-\frac{h r}{R} \sin\theta\right) + \frac{h}{R} r \cdot r \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} \cos^2\theta - \frac{h r^2}{R} \sin^2\theta + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} (\cos^2\theta + \sin^2\theta) + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} (1) + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \times 0 = 0\)
Then,
\(\displaystyle V = 0 + \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
Or
\(\displaystyle V = \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
Your task is now to find \(\displaystyle \bold{m}_2\).
\(\displaystyle \bold{m}_2 = \bold{r}_{2r} \times \bold{r}_{2\theta}\)
\(\displaystyle \mathbf{r}_{2r} = \frac{\partial \mathbf{r}_2}{\partial r} = \langle \cos\theta, \ \sin\theta, \ 0 \rangle\)
\(\displaystyle \mathbf{r}_{2\theta} = \frac{\partial \mathbf{r}_2}{\partial \theta} = \langle -r \sin\theta, \ r \cos\theta, \ 0 \rangle\)
Then,
\(\displaystyle \mathbf{m}_2 = \mathbf{r}_{2r} \times \mathbf{r}_{2\theta} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & 0 \\-r \sin\theta & r \cos\theta & 0\end{vmatrix} = \langle 0,\ 0,\ r \rangle\)
Your task now is to take the dot product.
\(\displaystyle \mathbf{F} \cdot \mathbf{m}_2 = \frac{1}{3} \langle r \cos\theta,\, r \sin\theta,\, h \rangle \cdot \langle 0,\, 0,\, r \rangle\)
\(\displaystyle = \frac{1}{3} \left( r \cos\theta \cdot 0 + r \sin\theta \cdot 0 + h \cdot r \right)\)
\(\displaystyle = \frac{1}{3} h r\)
Then,
\(\displaystyle V = \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta = \iint_{R_2} \frac{1}{3} h r \ dr \ d\theta\)
Or
\(\displaystyle V = \int_{0}^{2\pi} \int_{0}^{r} \frac{1}{3} h r \ dr \ d\theta = \int_{0}^{2\pi} \frac{1}{3} h \frac{r^2}{2} \ d\theta = 2\pi\frac{1}{3} h \frac{r^2}{2}\)
Or
\(\displaystyle V = \textcolor{indigo}{\frac{\pi r^2 h}{3}}\)
You sent your result to Khan and immediately received a message on your phone saying your bank account was credited with \(\displaystyle \textcolor{blue}{\$22,681.36}\).
Suppose that you got up in the morning bored and broke. You have no money and coincidently today is your wife's birthday and a month ago you promised yourself to buy her an expensive birthday present. Suddenly, you got a call from your Indian rich friend @khansaheb. You answered him and he has noticed that you are not okay from the tone of your voice. You explained to him your current situation and he promised to give you \(\displaystyle 2\)-million Indian rupees if you could solve the super integral to find out the volume of the cone in the OP. That is, you need to solve this:
\(\displaystyle V = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n} \ dS \quad \text{where} \quad \mathbf{F} = \frac{1}{3} \langle x, y, z \rangle\)
You have dug deep in your Calculus \(\displaystyle \text{III}\) notes, and you found out that the volume can be written as:
\(\displaystyle V = \iiint dV\)
Therefore, your first task was to prove:
\(\displaystyle \nabla \cdot \mathbf{F} = 1\)
Let \(\displaystyle \mathbf{F} = \langle F_1, F_2, F_3 \rangle = \frac{1}{3} \langle x, y, z \rangle\)
Then,
\(\displaystyle \nabla \cdot \mathbf{F} = \frac{1}{3}\left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\right) = \frac{1}{3}\bigg(1 + 1 + 1\bigg) = 1\)
Your first task has been done successfully. Now you can move to the second task which is to prove:
\(\displaystyle \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \frac{\pi r^2 h}{3}\)
You were very good in Calculus \(\displaystyle \text{III}\) and instantly you noticed that the cone can be described by this equation:
\(\displaystyle z = \frac{h}{R}\sqrt{x^2 + y^2}\)
Or
\(\displaystyle f(x,y) = z = \frac{h}{R}\sqrt{x^2 + y^2}\)
Then, the position vector is:
\(\displaystyle \bold{r} = \langle x, y, f(x,y) \rangle\)
And
\(\displaystyle \bold{n} = \frac{\bold{r}_x \times \bold{r}_y}{|\bold{r}_x \times \bold{r}_y|}\)
\(\displaystyle dS = |\bold{r}_x \times \bold{r}_y| \ dx \ dy\)
Let \(\displaystyle \bold{m}\) be the normal vector, then
\(\displaystyle \bold{m} = \bold{r}_x \times \bold{r}_y\)
Or
\(\displaystyle \bold{n} \ dS = \frac{\bold{m}}{|\bold{r}_x \times \bold{r}_y|}|\bold{r}_x \times \bold{r}_y| \ dx \ dy = \bold{m} \ dx \ dy\)
Then,
\(\displaystyle \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \iint_R \mathbf{F} \cdot \bold{m} \ dx \ dy\)
Your wife made you an orange juice and while you were drinking you noticed that:
\(\displaystyle z = \frac{h}{R}\sqrt{x^2 + y^2} = \frac{h}{R}r\)
You immediately called Khan and asked him if he may allow you to work in the Cylindrical coordinate as from your long experience you know that the Cartesian coordinate is not always friendly in solving integrals. Khan as a gentleman accepted.
Therefore, your integral becomes:
\(\displaystyle \iint_R \mathbf{F} \cdot \bold{m} \ dx \ dy = \iint_R \mathbf{F} \cdot \bold{m} \ dr \ d\theta\)
where the position vector is now:
\(\displaystyle \bold{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, z \rangle\)
So that \(\displaystyle \bold{m} = \bold{r}_{r} \times \bold{r}_{\theta}\)
You are very smart, so you noticed that the cone has two surface areas. One is the slant surface area while the other is the base of the cone.
Therefore, the integral Khan gave you becomes:
\(\displaystyle V = \iint_S \mathbf{F} \cdot \mathbf{n} \ dS = \iint_{R_1} \mathbf{F} \cdot \bold{m}_1 \ dr \ d\theta + \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
This means you have to work with two different position vectors:
\(\displaystyle \bold{r}_1(r,\theta) = \left\langle r\cos\theta, r\sin\theta, \frac{h}{R}r \right\rangle\)
\(\displaystyle \bold{r}_2(r,\theta) = \langle r\cos\theta, r\sin\theta, h \rangle\)
You decided to work with the first integral first:
\(\displaystyle V_1 = \iint_{R_1} \mathbf{F} \cdot \bold{m}_1 \ dr \ d\theta\)
Your task is now to find \(\displaystyle \bold{m}_1\).
\(\displaystyle \bold{m}_1 = \bold{r}_{1r} \times \bold{r}_{1\theta}\)
\(\displaystyle \bold{r}_{1r} = \frac{\partial \bold{r}_1}{\partial r} = \left\langle \cos\theta, \sin\theta, \frac{h}{R} \right\rangle\)
\(\displaystyle \bold{r}_{1\theta} = \frac{\partial \bold{r}_1}{\partial \theta} = \left\langle -r \sin\theta, r \cos\theta, 0 \right\rangle\)
Then,
\(\displaystyle \bold{m}_1 = \bold{r}_{1r} \times \bold{r}_{1\theta} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & \frac{h}{R} \\-r \sin\theta & r \cos\theta & 0\end{vmatrix} = \left\langle -\frac{h r}{R} \cos\theta, -\frac{h r}{R} \sin\theta, r \right\rangle\)
Now your task is to take the dot product. That is:
\(\displaystyle \bold{F} = \frac{1}{3} \langle x, y, z \rangle = \frac{1}{3} \langle r \cos\theta, r \sin\theta, \frac{h}{R} r \rangle\)
\(\displaystyle \bold{F} \cdot \bold{m}_1 = \frac{1}{3} \langle r \cos\theta, r \sin\theta, \frac{h}{R} r \rangle \cdot \left\langle -\frac{h r}{R} \cos\theta, -\frac{h r}{R} \sin\theta, r \right\rangle\)
\(\displaystyle = \frac{1}{3} \left( r \cos\theta \cdot \left(-\frac{h r}{R} \cos\theta\right) + r \sin\theta \cdot \left(-\frac{h r}{R} \sin\theta\right) + \frac{h}{R} r \cdot r \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} \cos^2\theta - \frac{h r^2}{R} \sin^2\theta + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} (\cos^2\theta + \sin^2\theta) + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \left( -\frac{h r^2}{R} (1) + \frac{h r^2}{R} \right)\)
\(\displaystyle = \frac{1}{3} \times 0 = 0\)
Then,
\(\displaystyle V = 0 + \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
Or
\(\displaystyle V = \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta\)
Your task is now to find \(\displaystyle \bold{m}_2\).
\(\displaystyle \bold{m}_2 = \bold{r}_{2r} \times \bold{r}_{2\theta}\)
\(\displaystyle \mathbf{r}_{2r} = \frac{\partial \mathbf{r}_2}{\partial r} = \langle \cos\theta, \ \sin\theta, \ 0 \rangle\)
\(\displaystyle \mathbf{r}_{2\theta} = \frac{\partial \mathbf{r}_2}{\partial \theta} = \langle -r \sin\theta, \ r \cos\theta, \ 0 \rangle\)
Then,
\(\displaystyle \mathbf{m}_2 = \mathbf{r}_{2r} \times \mathbf{r}_{2\theta} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & 0 \\-r \sin\theta & r \cos\theta & 0\end{vmatrix} = \langle 0,\ 0,\ r \rangle\)
Your task now is to take the dot product.
\(\displaystyle \mathbf{F} \cdot \mathbf{m}_2 = \frac{1}{3} \langle r \cos\theta,\, r \sin\theta,\, h \rangle \cdot \langle 0,\, 0,\, r \rangle\)
\(\displaystyle = \frac{1}{3} \left( r \cos\theta \cdot 0 + r \sin\theta \cdot 0 + h \cdot r \right)\)
\(\displaystyle = \frac{1}{3} h r\)
Then,
\(\displaystyle V = \iint_{R_2} \mathbf{F} \cdot \bold{m}_2 \ dr \ d\theta = \iint_{R_2} \frac{1}{3} h r \ dr \ d\theta\)
Or
\(\displaystyle V = \int_{0}^{2\pi} \int_{0}^{r} \frac{1}{3} h r \ dr \ d\theta = \int_{0}^{2\pi} \frac{1}{3} h \frac{r^2}{2} \ d\theta = 2\pi\frac{1}{3} h \frac{r^2}{2}\)
Or
\(\displaystyle V = \textcolor{indigo}{\frac{\pi r^2 h}{3}}\)
You sent your result to Khan and immediately received a message on your phone saying your bank account was credited with \(\displaystyle \textcolor{blue}{\$22,681.36}\).
Agent Smith
Full Member
- Joined
- Oct 18, 2023
- Messages
- 530
Ok sir, but I don't think khansaheb thinks of me as a friend. If this site were an inn, we just happened to lock gazes while dining. He smiles at you and I smile at him smiling at you.
Good job though, with your calculations. Please post more questions and their answers. It'll keep you sharp and us informed. Now who wouldn't give an arm and a leg for this golden opportunity?
Good job though, with your calculations. Please post more questions and their answers. It'll keep you sharp and us informed. Now who wouldn't give an arm and a leg for this golden opportunity?