How to find the area of a cardioid using multivariable calculus?
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So I have the following three questions (and no solutions to them sadly). I wanted to know if my results for a) and b) are correct and how I should proceed for c).
Given is the following ellipsoid : $$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} leq 1$$ and the following questions :
a) Calculate the volume of the ellipsoid
b) Calculate the flux of $V=(x-3xy^2,y^3-2xyz,xz^2+2z)$ through the surface of the ellispoid toward the exterior
c) Given is the cardiode $(x^2+y^2-2x)^2=4(x^2+y^2)$. Determine the "area which is enclosed by the cardioid" (german: Flächeninhalt).
a) As usual, we make the substitutions $hat{x}=frac{x}{a}$, $hat{y}=frac{y}{b}$ and $hat{z}=frac{z}{c}$, we get $dx=adhat{x}, dy=bdhat{y}, dz = cdhat{z}$, so now we have a sphere and can set up our triple integral in spherical coordinates $$abc int_{0}^{1} int_{0}^{2pi}int_{0}^{pi}1*r^2sin(phi)dphi dtheta dr$$ which after integration yields the well known $abc frac{4pi}{3}$
b) The divergence of $V$ is simply 3. Thus we can use the divergence theorem : we have a triple integral (i.e. the volume of your ellipsoid) over the divergence of our vector field. In our case, the divergence doesn't depend on x, y or z, so we simply get our volume times our divergence, i.e. $abc frac{4pi}{3}*3=abc 4 pi$
c) Now, here I really have some troubles, because I'm supposed to calculate the "area enclosed" using multivariable calculus, i.e. not just applying a well known formula. How am I supposed to proceed ? Should I use polar coordinates or what ?
Thanks for your help !
Edit 11/01/2019:
In fact, c) can be solved really easily. If we just use polar coordinates $x=rcos(theta), y=rsin(theta)$, we get $(r^2-2rcos(theta))^2=4r^2$. If we take the square root, the right side has to be positive because $r$ cannot be negative, so we get $2r$. For the left side, we have $pm(r^2-2rcos(theta))=2r$ . So either $r(r-2cos(theta)-2)=0$ and either $r=0$ or $r=2(cos(theta)+1)$ or $r(-r+2cos(theta)-2)=0$ and $r=0$ or $r=2(cos(theta)-1)$ which is not possible because $r$ would be negative if not zero.
Thus $r=2(cos(theta)+1)$. Now we can set up our double integral:
$$int_{0}^{2pi}int_{0}^{2(cos(theta)+1)}1 r dr dtheta=int_{0}^{2pi}2(cos(theta)+1)^2 dtheta =2 int_{0}^{2pi}(cos^2(theta)+2cos(theta)+1) dtheta=2(pi+ 0 +2pi)=6pi$$
real-analysis integration analysis multivariable-calculus divergence
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
$endgroup$
|
show 1 more comment
$begingroup$
So I have the following three questions (and no solutions to them sadly). I wanted to know if my results for a) and b) are correct and how I should proceed for c).
Given is the following ellipsoid : $$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} leq 1$$ and the following questions :
a) Calculate the volume of the ellipsoid
b) Calculate the flux of $V=(x-3xy^2,y^3-2xyz,xz^2+2z)$ through the surface of the ellispoid toward the exterior
c) Given is the cardiode $(x^2+y^2-2x)^2=4(x^2+y^2)$. Determine the "area which is enclosed by the cardioid" (german: Flächeninhalt).
a) As usual, we make the substitutions $hat{x}=frac{x}{a}$, $hat{y}=frac{y}{b}$ and $hat{z}=frac{z}{c}$, we get $dx=adhat{x}, dy=bdhat{y}, dz = cdhat{z}$, so now we have a sphere and can set up our triple integral in spherical coordinates $$abc int_{0}^{1} int_{0}^{2pi}int_{0}^{pi}1*r^2sin(phi)dphi dtheta dr$$ which after integration yields the well known $abc frac{4pi}{3}$
b) The divergence of $V$ is simply 3. Thus we can use the divergence theorem : we have a triple integral (i.e. the volume of your ellipsoid) over the divergence of our vector field. In our case, the divergence doesn't depend on x, y or z, so we simply get our volume times our divergence, i.e. $abc frac{4pi}{3}*3=abc 4 pi$
c) Now, here I really have some troubles, because I'm supposed to calculate the "area enclosed" using multivariable calculus, i.e. not just applying a well known formula. How am I supposed to proceed ? Should I use polar coordinates or what ?
Thanks for your help !
Edit 11/01/2019:
In fact, c) can be solved really easily. If we just use polar coordinates $x=rcos(theta), y=rsin(theta)$, we get $(r^2-2rcos(theta))^2=4r^2$. If we take the square root, the right side has to be positive because $r$ cannot be negative, so we get $2r$. For the left side, we have $pm(r^2-2rcos(theta))=2r$ . So either $r(r-2cos(theta)-2)=0$ and either $r=0$ or $r=2(cos(theta)+1)$ or $r(-r+2cos(theta)-2)=0$ and $r=0$ or $r=2(cos(theta)-1)$ which is not possible because $r$ would be negative if not zero.
Thus $r=2(cos(theta)+1)$. Now we can set up our double integral:
$$int_{0}^{2pi}int_{0}^{2(cos(theta)+1)}1 r dr dtheta=int_{0}^{2pi}2(cos(theta)+1)^2 dtheta =2 int_{0}^{2pi}(cos^2(theta)+2cos(theta)+1) dtheta=2(pi+ 0 +2pi)=6pi$$
real-analysis integration analysis multivariable-calculus divergence
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
$endgroup$
$begingroup$
(c) yes, use polar coordinates. Let's see what you get
$endgroup$
– Lozenges
Dec 21 '18 at 13:10
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@Lozenges Ok, I will try
$endgroup$
– Poujh
Dec 21 '18 at 13:17
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Why the downvote ?
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– Poujh
Dec 21 '18 at 14:14
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(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
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– Jyrki Lahtonen
Jan 11 at 13:25
$begingroup$
@JyrkiLahtonen Ok, I will think about it next time. Thank you.
$endgroup$
– Poujh
Jan 11 at 13:56
|
show 1 more comment
$begingroup$
So I have the following three questions (and no solutions to them sadly). I wanted to know if my results for a) and b) are correct and how I should proceed for c).
Given is the following ellipsoid : $$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} leq 1$$ and the following questions :
a) Calculate the volume of the ellipsoid
b) Calculate the flux of $V=(x-3xy^2,y^3-2xyz,xz^2+2z)$ through the surface of the ellispoid toward the exterior
c) Given is the cardiode $(x^2+y^2-2x)^2=4(x^2+y^2)$. Determine the "area which is enclosed by the cardioid" (german: Flächeninhalt).
a) As usual, we make the substitutions $hat{x}=frac{x}{a}$, $hat{y}=frac{y}{b}$ and $hat{z}=frac{z}{c}$, we get $dx=adhat{x}, dy=bdhat{y}, dz = cdhat{z}$, so now we have a sphere and can set up our triple integral in spherical coordinates $$abc int_{0}^{1} int_{0}^{2pi}int_{0}^{pi}1*r^2sin(phi)dphi dtheta dr$$ which after integration yields the well known $abc frac{4pi}{3}$
b) The divergence of $V$ is simply 3. Thus we can use the divergence theorem : we have a triple integral (i.e. the volume of your ellipsoid) over the divergence of our vector field. In our case, the divergence doesn't depend on x, y or z, so we simply get our volume times our divergence, i.e. $abc frac{4pi}{3}*3=abc 4 pi$
c) Now, here I really have some troubles, because I'm supposed to calculate the "area enclosed" using multivariable calculus, i.e. not just applying a well known formula. How am I supposed to proceed ? Should I use polar coordinates or what ?
Thanks for your help !
Edit 11/01/2019:
In fact, c) can be solved really easily. If we just use polar coordinates $x=rcos(theta), y=rsin(theta)$, we get $(r^2-2rcos(theta))^2=4r^2$. If we take the square root, the right side has to be positive because $r$ cannot be negative, so we get $2r$. For the left side, we have $pm(r^2-2rcos(theta))=2r$ . So either $r(r-2cos(theta)-2)=0$ and either $r=0$ or $r=2(cos(theta)+1)$ or $r(-r+2cos(theta)-2)=0$ and $r=0$ or $r=2(cos(theta)-1)$ which is not possible because $r$ would be negative if not zero.
Thus $r=2(cos(theta)+1)$. Now we can set up our double integral:
$$int_{0}^{2pi}int_{0}^{2(cos(theta)+1)}1 r dr dtheta=int_{0}^{2pi}2(cos(theta)+1)^2 dtheta =2 int_{0}^{2pi}(cos^2(theta)+2cos(theta)+1) dtheta=2(pi+ 0 +2pi)=6pi$$
real-analysis integration analysis multivariable-calculus divergence
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
$endgroup$
So I have the following three questions (and no solutions to them sadly). I wanted to know if my results for a) and b) are correct and how I should proceed for c).
Given is the following ellipsoid : $$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} leq 1$$ and the following questions :
a) Calculate the volume of the ellipsoid
b) Calculate the flux of $V=(x-3xy^2,y^3-2xyz,xz^2+2z)$ through the surface of the ellispoid toward the exterior
c) Given is the cardiode $(x^2+y^2-2x)^2=4(x^2+y^2)$. Determine the "area which is enclosed by the cardioid" (german: Flächeninhalt).
a) As usual, we make the substitutions $hat{x}=frac{x}{a}$, $hat{y}=frac{y}{b}$ and $hat{z}=frac{z}{c}$, we get $dx=adhat{x}, dy=bdhat{y}, dz = cdhat{z}$, so now we have a sphere and can set up our triple integral in spherical coordinates $$abc int_{0}^{1} int_{0}^{2pi}int_{0}^{pi}1*r^2sin(phi)dphi dtheta dr$$ which after integration yields the well known $abc frac{4pi}{3}$
b) The divergence of $V$ is simply 3. Thus we can use the divergence theorem : we have a triple integral (i.e. the volume of your ellipsoid) over the divergence of our vector field. In our case, the divergence doesn't depend on x, y or z, so we simply get our volume times our divergence, i.e. $abc frac{4pi}{3}*3=abc 4 pi$
c) Now, here I really have some troubles, because I'm supposed to calculate the "area enclosed" using multivariable calculus, i.e. not just applying a well known formula. How am I supposed to proceed ? Should I use polar coordinates or what ?
Thanks for your help !
Edit 11/01/2019:
In fact, c) can be solved really easily. If we just use polar coordinates $x=rcos(theta), y=rsin(theta)$, we get $(r^2-2rcos(theta))^2=4r^2$. If we take the square root, the right side has to be positive because $r$ cannot be negative, so we get $2r$. For the left side, we have $pm(r^2-2rcos(theta))=2r$ . So either $r(r-2cos(theta)-2)=0$ and either $r=0$ or $r=2(cos(theta)+1)$ or $r(-r+2cos(theta)-2)=0$ and $r=0$ or $r=2(cos(theta)-1)$ which is not possible because $r$ would be negative if not zero.
Thus $r=2(cos(theta)+1)$. Now we can set up our double integral:
$$int_{0}^{2pi}int_{0}^{2(cos(theta)+1)}1 r dr dtheta=int_{0}^{2pi}2(cos(theta)+1)^2 dtheta =2 int_{0}^{2pi}(cos^2(theta)+2cos(theta)+1) dtheta=2(pi+ 0 +2pi)=6pi$$
real-analysis integration analysis multivariable-calculus divergence
real-analysis integration analysis multivariable-calculus divergence
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
edited Jan 11 at 13:04
Poujh
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
asked Dec 21 '18 at 12:33
PoujhPoujh
6111516
6111516
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(c) yes, use polar coordinates. Let's see what you get
$endgroup$
– Lozenges
Dec 21 '18 at 13:10
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@Lozenges Ok, I will try
$endgroup$
– Poujh
Dec 21 '18 at 13:17
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Why the downvote ?
$endgroup$
– Poujh
Dec 21 '18 at 14:14
$begingroup$
(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 13:25
$begingroup$
@JyrkiLahtonen Ok, I will think about it next time. Thank you.
$endgroup$
– Poujh
Jan 11 at 13:56
|
show 1 more comment
$begingroup$
(c) yes, use polar coordinates. Let's see what you get
$endgroup$
– Lozenges
Dec 21 '18 at 13:10
$begingroup$
@Lozenges Ok, I will try
$endgroup$
– Poujh
Dec 21 '18 at 13:17
$begingroup$
Why the downvote ?
$endgroup$
– Poujh
Dec 21 '18 at 14:14
$begingroup$
(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 13:25
$begingroup$
@JyrkiLahtonen Ok, I will think about it next time. Thank you.
$endgroup$
– Poujh
Jan 11 at 13:56
$begingroup$
(c) yes, use polar coordinates. Let's see what you get
$endgroup$
– Lozenges
Dec 21 '18 at 13:10
$begingroup$
(c) yes, use polar coordinates. Let's see what you get
$endgroup$
– Lozenges
Dec 21 '18 at 13:10
$begingroup$
@Lozenges Ok, I will try
$endgroup$
– Poujh
Dec 21 '18 at 13:17
$begingroup$
@Lozenges Ok, I will try
$endgroup$
– Poujh
Dec 21 '18 at 13:17
$begingroup$
Why the downvote ?
$endgroup$
– Poujh
Dec 21 '18 at 14:14
$begingroup$
Why the downvote ?
$endgroup$
– Poujh
Dec 21 '18 at 14:14
$begingroup$
(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 13:25
$begingroup$
(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 13:25
$begingroup$
@JyrkiLahtonen Ok, I will think about it next time. Thank you.
$endgroup$
– Poujh
Jan 11 at 13:56
$begingroup$
@JyrkiLahtonen Ok, I will think about it next time. Thank you.
$endgroup$
– Poujh
Jan 11 at 13:56
|
show 1 more comment
1 Answer
1
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- Write the equation in polar coordinates, and try to draw what it is.
- Construct a simple vector field whose divergence is identically 1, and then use the divergence theorem (or Green's theorem) to reduce the area into an integral over the curve.
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
$endgroup$
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
$endgroup$
– Poujh
Dec 21 '18 at 14:06
$begingroup$
You have a formula for computing area in polar coordinates. Use that formula.
$endgroup$
– timur
Dec 21 '18 at 14:07
|
show 6 more comments
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1 Answer
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1 Answer
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$begingroup$
- Write the equation in polar coordinates, and try to draw what it is.
- Construct a simple vector field whose divergence is identically 1, and then use the divergence theorem (or Green's theorem) to reduce the area into an integral over the curve.
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
$endgroup$
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
$endgroup$
– Poujh
Dec 21 '18 at 14:06
$begingroup$
You have a formula for computing area in polar coordinates. Use that formula.
$endgroup$
– timur
Dec 21 '18 at 14:07
|
show 6 more comments
$begingroup$
- Write the equation in polar coordinates, and try to draw what it is.
- Construct a simple vector field whose divergence is identically 1, and then use the divergence theorem (or Green's theorem) to reduce the area into an integral over the curve.
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
$endgroup$
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
$endgroup$
– Poujh
Dec 21 '18 at 14:06
$begingroup$
You have a formula for computing area in polar coordinates. Use that formula.
$endgroup$
– timur
Dec 21 '18 at 14:07
|
show 6 more comments
$begingroup$
- Write the equation in polar coordinates, and try to draw what it is.
- Construct a simple vector field whose divergence is identically 1, and then use the divergence theorem (or Green's theorem) to reduce the area into an integral over the curve.
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
$endgroup$
- Write the equation in polar coordinates, and try to draw what it is.
- Construct a simple vector field whose divergence is identically 1, and then use the divergence theorem (or Green's theorem) to reduce the area into an integral over the curve.
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
answered Dec 21 '18 at 13:18
timurtimur
12.2k2144
12.2k2144
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
$endgroup$
– Poujh
Dec 21 '18 at 14:06
$begingroup$
You have a formula for computing area in polar coordinates. Use that formula.
$endgroup$
– timur
Dec 21 '18 at 14:07
|
show 6 more comments
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
$endgroup$
– Poujh
Dec 21 '18 at 14:06
$begingroup$
You have a formula for computing area in polar coordinates. Use that formula.
$endgroup$
– timur
Dec 21 '18 at 14:07
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
$begingroup$
Well, we haven't covered Green's theorem yet, so I don't think I am allowed to use it.
$endgroup$
– Poujh
Dec 21 '18 at 13:22
1
1
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
@Poujh Then it is polar
$endgroup$
– timur
Dec 21 '18 at 14:02
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
I did it wrong. I should get $2int_{0}^{2pi}1-2cos(theta)+cos^2(theta)$ which evaluates to $6pi$, which seems to be correct given that I also obtain $6pi$ using the area formula $frac{3}{2}pi a^2$ where $a=-2$
$endgroup$
– Poujh
Dec 21 '18 at 14:04
$begingroup$
What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
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– Poujh
Dec 21 '18 at 14:06
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What do you mean exactly with "polar" ? And just to be clear, is what I just said above all I need to do ? I mean, just use $int_{0}^{2pi}frac{1}{2}r^2$ where $r=-2(1-cos(theta))$ ?
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– Poujh
Dec 21 '18 at 14:06
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You have a formula for computing area in polar coordinates. Use that formula.
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– timur
Dec 21 '18 at 14:07
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You have a formula for computing area in polar coordinates. Use that formula.
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– timur
Dec 21 '18 at 14:07
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(c) yes, use polar coordinates. Let's see what you get
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– Lozenges
Dec 21 '18 at 13:10
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@Lozenges Ok, I will try
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– Poujh
Dec 21 '18 at 13:17
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Why the downvote ?
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– Poujh
Dec 21 '18 at 14:14
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(c) is largely unrelated to (a) and (b). You should ask one question per ... question.
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– Jyrki Lahtonen
Jan 11 at 13:25
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@JyrkiLahtonen Ok, I will think about it next time. Thank you.
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– Poujh
Jan 11 at 13:56