Mixing
How does one find the common area shared between two polar curves?
$begingroup$
The problem: Consider the circle $r = (sqrt{2} - 1)cos(theta)$ and the cardioid $r = 1 - cos(theta)$. Find the area $A$ of the region $R$ common to the inside of the circle and the inside of the cardioid.
I am confused about solving this. Can someone in a not too technical/difficult way explain how this would be solved?
Thank you
polar-coordinates
asked Dec 14 '16 at 3:31
junjun
64
$endgroup$
add a comment |
$begingroup$
The problem: Consider the circle $r = (sqrt{2} - 1)cos(theta)$ and the cardioid $r = 1 - cos(theta)$. Find the area $A$ of the region $R$ common to the inside of the circle and the inside of the cardioid.
I am confused about solving this. Can someone in a not too technical/difficult way explain how this would be solved?
Thank you
polar-coordinates
asked Dec 14 '16 at 3:31
junjun
64
$endgroup$
add a comment |
$begingroup$
The problem: Consider the circle $r = (sqrt{2} - 1)cos(theta)$ and the cardioid $r = 1 - cos(theta)$. Find the area $A$ of the region $R$ common to the inside of the circle and the inside of the cardioid.
I am confused about solving this. Can someone in a not too technical/difficult way explain how this would be solved?
Thank you
polar-coordinates
asked Dec 14 '16 at 3:31
junjun
64
$endgroup$
The problem: Consider the circle $r = (sqrt{2} - 1)cos(theta)$ and the cardioid $r = 1 - cos(theta)$. Find the area $A$ of the region $R$ common to the inside of the circle and the inside of the cardioid.
I am confused about solving this. Can someone in a not too technical/difficult way explain how this would be solved?
Thank you
polar-coordinates
polar-coordinates
asked Dec 14 '16 at 3:31
junjun
64
asked Dec 14 '16 at 3:31
junjun
64
asked Dec 14 '16 at 3:31
junjun
64
asked Dec 14 '16 at 3:31
junjun
64
asked Dec 14 '16 at 3:31
junjun
64
64
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Firstly, let's consider the graphs involved:
Now you can at least see the area that is being asked about. (It's the two little wingy-dingys, one to the upper right and one to the lower right.)
To solve this, you will need to use calculus; specifically, integral calculus.
The basic idea of integral calculus is adding up a lot of small, tiny pieces to get the measure of the whole.
With Cartesian coordinates ($x$ and $y$), you add up tiny little rectangles. Since these are polar coordinates, you should add up tiny little sectors.
How to do this precisely is the subject of a first-year calculus course, but this is the underlying theory.
And, from just eyeballing the above graph, you can sanity-check your exact answer.
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
$endgroup$
add a comment |
$begingroup$
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(sqrt2-1)costheta=1-costheta$ which holds when $theta=pmpi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $pm45$ degree angles away from the origin.
Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $thetain[0,pi/4]$ together with the part of the circle in the interval $thetain[pi/4,pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$
A=frac12int_0^{pi/4}(1-cos theta)^2,dtheta+frac12int_{pi/4}^{pi/2}(sqrt2-1)^2cos^2theta,dtheta.
$$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $thetain[-pi/4,pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$
frac12int_{-pi/4}^{pi/4}left((sqrt2-1)^2cos^2theta-(1-costheta)^2right),dtheta$$
instead.
$$
Moral: Do not forget to draw the picture.
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Firstly, let's consider the graphs involved:
Now you can at least see the area that is being asked about. (It's the two little wingy-dingys, one to the upper right and one to the lower right.)
To solve this, you will need to use calculus; specifically, integral calculus.
The basic idea of integral calculus is adding up a lot of small, tiny pieces to get the measure of the whole.
With Cartesian coordinates ($x$ and $y$), you add up tiny little rectangles. Since these are polar coordinates, you should add up tiny little sectors.
How to do this precisely is the subject of a first-year calculus course, but this is the underlying theory.
And, from just eyeballing the above graph, you can sanity-check your exact answer.
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
$endgroup$
add a comment |
$begingroup$
Firstly, let's consider the graphs involved:
Now you can at least see the area that is being asked about. (It's the two little wingy-dingys, one to the upper right and one to the lower right.)
To solve this, you will need to use calculus; specifically, integral calculus.
The basic idea of integral calculus is adding up a lot of small, tiny pieces to get the measure of the whole.
With Cartesian coordinates ($x$ and $y$), you add up tiny little rectangles. Since these are polar coordinates, you should add up tiny little sectors.
How to do this precisely is the subject of a first-year calculus course, but this is the underlying theory.
And, from just eyeballing the above graph, you can sanity-check your exact answer.
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
$endgroup$
add a comment |
$begingroup$
Firstly, let's consider the graphs involved:
Now you can at least see the area that is being asked about. (It's the two little wingy-dingys, one to the upper right and one to the lower right.)
To solve this, you will need to use calculus; specifically, integral calculus.
The basic idea of integral calculus is adding up a lot of small, tiny pieces to get the measure of the whole.
With Cartesian coordinates ($x$ and $y$), you add up tiny little rectangles. Since these are polar coordinates, you should add up tiny little sectors.
How to do this precisely is the subject of a first-year calculus course, but this is the underlying theory.
And, from just eyeballing the above graph, you can sanity-check your exact answer.
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
$endgroup$
Firstly, let's consider the graphs involved:
Now you can at least see the area that is being asked about. (It's the two little wingy-dingys, one to the upper right and one to the lower right.)
To solve this, you will need to use calculus; specifically, integral calculus.
The basic idea of integral calculus is adding up a lot of small, tiny pieces to get the measure of the whole.
With Cartesian coordinates ($x$ and $y$), you add up tiny little rectangles. Since these are polar coordinates, you should add up tiny little sectors.
How to do this precisely is the subject of a first-year calculus course, but this is the underlying theory.
And, from just eyeballing the above graph, you can sanity-check your exact answer.
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
answered Dec 14 '16 at 4:31
WildcardWildcard
2,6251028
2,6251028
add a comment |
add a comment |
$begingroup$
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(sqrt2-1)costheta=1-costheta$ which holds when $theta=pmpi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $pm45$ degree angles away from the origin.
Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $thetain[0,pi/4]$ together with the part of the circle in the interval $thetain[pi/4,pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$
A=frac12int_0^{pi/4}(1-cos theta)^2,dtheta+frac12int_{pi/4}^{pi/2}(sqrt2-1)^2cos^2theta,dtheta.
$$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $thetain[-pi/4,pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$
frac12int_{-pi/4}^{pi/4}left((sqrt2-1)^2cos^2theta-(1-costheta)^2right),dtheta$$
instead.
$$
Moral: Do not forget to draw the picture.
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
$endgroup$
add a comment |
$begingroup$
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(sqrt2-1)costheta=1-costheta$ which holds when $theta=pmpi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $pm45$ degree angles away from the origin.
Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $thetain[0,pi/4]$ together with the part of the circle in the interval $thetain[pi/4,pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$
A=frac12int_0^{pi/4}(1-cos theta)^2,dtheta+frac12int_{pi/4}^{pi/2}(sqrt2-1)^2cos^2theta,dtheta.
$$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $thetain[-pi/4,pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$
frac12int_{-pi/4}^{pi/4}left((sqrt2-1)^2cos^2theta-(1-costheta)^2right),dtheta$$
instead.
$$
Moral: Do not forget to draw the picture.
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
$endgroup$
add a comment |
$begingroup$
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(sqrt2-1)costheta=1-costheta$ which holds when $theta=pmpi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $pm45$ degree angles away from the origin.
Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $thetain[0,pi/4]$ together with the part of the circle in the interval $thetain[pi/4,pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$
A=frac12int_0^{pi/4}(1-cos theta)^2,dtheta+frac12int_{pi/4}^{pi/2}(sqrt2-1)^2cos^2theta,dtheta.
$$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $thetain[-pi/4,pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$
frac12int_{-pi/4}^{pi/4}left((sqrt2-1)^2cos^2theta-(1-costheta)^2right),dtheta$$
instead.
$$
Moral: Do not forget to draw the picture.
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
$endgroup$
Before you can figure out what kind of an integral will give you the answer you should (at least mentally) sketch a picture of the curves.
Here is a picture with the entire cardioid and most of the circle shown:
You see that the two curves intersect at the origin and also at two other points symmetric about the $x$-axis. Those two points can be found by solving the equation $(sqrt2-1)costheta=1-costheta$ which holds when $theta=pmpi/4$.
Anyway, we see that the common region consists of those two lense shaped blobs "pointing" at $pm45$ degree angles away from the origin.
Here is another picture with all of the circle and essential parts of the cardioid.
Because of the symmetry it suffices to calculate the area of one of those lenses. We see that the boundary of that region consists of the part of the cardioid in the interval $thetain[0,pi/4]$ together with the part of the circle in the interval $thetain[pi/4,pi/2]$. Also observe that radially we are including everything between those two arcs and the origin. Therefore the lense has area
$$
A=frac12int_0^{pi/4}(1-cos theta)^2,dtheta+frac12int_{pi/4}^{pi/2}(sqrt2-1)^2cos^2theta,dtheta.
$$
Leaving the calculation of those integrals to you. Remember to multiply by two at the end to account for the other lense.
Observe that if the task were to calculate the region inside the circle but outside the cardioid, then the regions bounded by the two curves would radially overlap each other. That region is in the sector $thetain[-pi/4,pi/4]$, and within that sector the circle gives the outer perimeter and the cardioid the inner, so you would calculate the integral
$$
frac12int_{-pi/4}^{pi/4}left((sqrt2-1)^2cos^2theta-(1-costheta)^2right),dtheta$$
instead.
$$
Moral: Do not forget to draw the picture.
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
edited Dec 14 '16 at 4:36
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
answered Dec 14 '16 at 4:31
Jyrki LahtonenJyrki Lahtonen
109k13170376
109k13170376
add a comment |
add a comment |
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