Finding Bounds for Area Between Polar Curves Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Area Bounded by Polar CurvesFinding area between two polar curves using double integralsarea between two polar curvesCan someone check my answer for this area between 2 polar curves question?Area between two polar curves $r = 2 sintheta$ and $r =2costheta$Area of two polar regionsArea inside polar curveDetermining bounds for polar areaArea between polar curvesDetermining the area between two polar curves using a double integral
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Finding Bounds for Area Between Polar Curves
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Area Bounded by Polar CurvesFinding area between two polar curves using double integralsarea between two polar curvesCan someone check my answer for this area between 2 polar curves question?Area between two polar curves $r = 2 sintheta$ and $r =2costheta$Area of two polar regionsArea inside polar curveDetermining bounds for polar areaArea between polar curvesDetermining the area between two polar curves using a double integral
$begingroup$
I'm trying to find the area of the region both inside the circle $r= sinθ$ and outside the circle $r=√3 cosθ$ (both equations are in polar coordinates). Here is what it looks like: 
The two graphs intersect at the origin and the polar point $(r, theta) = (pi/3, fracsqrt32)$.
I thought the obvious answer would be to use the formula $A = frac12int_theta_1^theta_2 [R^2 - r^2] dtheta$ and integrate from $0$ to $pi/3$ as seen below:
$$A = frac12int_0^pi/3 [(sintheta)^2 - (sqrt3costheta)^2] dtheta$$
however, this is wrong. What they have instead is:
$$A = frac12int_pi/3^pi (sintheta)^2 dtheta quad - quad frac12int_pi/3^pi/2(sqrt3costheta)^2] dtheta$$
Can someone please explain how they got the bounds in their answer? And if there is a simpler way, for example, a slight alteration to the way I initially tried to do it, please let me know.
Thank you!
calculus area polar-coordinates
asked Mar 31 at 23:26
CodingMeeCodingMee
756
$endgroup$
add a comment |
$begingroup$
I'm trying to find the area of the region both inside the circle $r= sinθ$ and outside the circle $r=√3 cosθ$ (both equations are in polar coordinates). Here is what it looks like:
The two graphs intersect at the origin and the polar point $(r, theta) = (pi/3, fracsqrt32)$.
I thought the obvious answer would be to use the formula $A = frac12int_theta_1^theta_2 [R^2 - r^2] dtheta$ and integrate from $0$ to $pi/3$ as seen below:
$$A = frac12int_0^pi/3 [(sintheta)^2 - (sqrt3costheta)^2] dtheta$$
however, this is wrong. What they have instead is:
$$A = frac12int_pi/3^pi (sintheta)^2 dtheta quad - quad frac12int_pi/3^pi/2(sqrt3costheta)^2] dtheta$$
Can someone please explain how they got the bounds in their answer? And if there is a simpler way, for example, a slight alteration to the way I initially tried to do it, please let me know.
Thank you!
calculus area polar-coordinates
asked Mar 31 at 23:26
CodingMeeCodingMee
756
$endgroup$
$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58
add a comment |
$begingroup$
I'm trying to find the area of the region both inside the circle $r= sinθ$ and outside the circle $r=√3 cosθ$ (both equations are in polar coordinates). Here is what it looks like:
The two graphs intersect at the origin and the polar point $(r, theta) = (pi/3, fracsqrt32)$.
I thought the obvious answer would be to use the formula $A = frac12int_theta_1^theta_2 [R^2 - r^2] dtheta$ and integrate from $0$ to $pi/3$ as seen below:
$$A = frac12int_0^pi/3 [(sintheta)^2 - (sqrt3costheta)^2] dtheta$$
however, this is wrong. What they have instead is:
$$A = frac12int_pi/3^pi (sintheta)^2 dtheta quad - quad frac12int_pi/3^pi/2(sqrt3costheta)^2] dtheta$$
Can someone please explain how they got the bounds in their answer? And if there is a simpler way, for example, a slight alteration to the way I initially tried to do it, please let me know.
Thank you!
calculus area polar-coordinates
asked Mar 31 at 23:26
CodingMeeCodingMee
756
$endgroup$
I'm trying to find the area of the region both inside the circle $r= sinθ$ and outside the circle $r=√3 cosθ$ (both equations are in polar coordinates). Here is what it looks like:
The two graphs intersect at the origin and the polar point $(r, theta) = (pi/3, fracsqrt32)$.
I thought the obvious answer would be to use the formula $A = frac12int_theta_1^theta_2 [R^2 - r^2] dtheta$ and integrate from $0$ to $pi/3$ as seen below:
$$A = frac12int_0^pi/3 [(sintheta)^2 - (sqrt3costheta)^2] dtheta$$
however, this is wrong. What they have instead is:
$$A = frac12int_pi/3^pi (sintheta)^2 dtheta quad - quad frac12int_pi/3^pi/2(sqrt3costheta)^2] dtheta$$
Can someone please explain how they got the bounds in their answer? And if there is a simpler way, for example, a slight alteration to the way I initially tried to do it, please let me know.
Thank you!
calculus area polar-coordinates
calculus area polar-coordinates
asked Mar 31 at 23:26
CodingMeeCodingMee
756
asked Mar 31 at 23:26
CodingMeeCodingMee
756
asked Mar 31 at 23:26
CodingMeeCodingMee
756
asked Mar 31 at 23:26
CodingMeeCodingMee
756
asked Mar 31 at 23:26
CodingMeeCodingMee
756
756
$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58
add a comment |
$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58
$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 le theta le pi$.
You can see that when $0 le theta < pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $pi/3 le theta le pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = sin theta$ and outside the circle $r = sqrt3 cos theta$. However, an important subtlety occurs when $theta = pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $pi/2 < theta < pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$frac12int_theta = pi/3^pi/2 ((sin theta)^2 - (sqrt3 cos theta)^2) , dtheta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$frac12 int_theta = pi/2^pi (sin theta)^2 , dtheta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.
answered Apr 1 at 4:14
heropupheropup
65.6k865104
$endgroup$
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
add a comment |
$begingroup$
The intersection for $x$ correspond to
$$sin(theta)cos(theta)=sqrt 3 cos(theta)cos(theta)$$
$$cos(theta)big(sin(theta)-sqrt 3 cos(theta)big)=0$$ then the two solutions corresponding to $cos(theta)=0$ and $sin(theta)-sqrt 3 cos(theta)=0$
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
oldest
votes
$begingroup$
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 le theta le pi$.
You can see that when $0 le theta < pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $pi/3 le theta le pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = sin theta$ and outside the circle $r = sqrt3 cos theta$. However, an important subtlety occurs when $theta = pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $pi/2 < theta < pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$frac12int_theta = pi/3^pi/2 ((sin theta)^2 - (sqrt3 cos theta)^2) , dtheta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$frac12 int_theta = pi/2^pi (sin theta)^2 , dtheta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.
answered Apr 1 at 4:14
heropupheropup
65.6k865104
$endgroup$
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
add a comment |
$begingroup$
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 le theta le pi$.
You can see that when $0 le theta < pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $pi/3 le theta le pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = sin theta$ and outside the circle $r = sqrt3 cos theta$. However, an important subtlety occurs when $theta = pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $pi/2 < theta < pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$frac12int_theta = pi/3^pi/2 ((sin theta)^2 - (sqrt3 cos theta)^2) , dtheta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$frac12 int_theta = pi/2^pi (sin theta)^2 , dtheta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.
answered Apr 1 at 4:14
heropupheropup
65.6k865104
$endgroup$
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
add a comment |
$begingroup$
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 le theta le pi$.
You can see that when $0 le theta < pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $pi/3 le theta le pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = sin theta$ and outside the circle $r = sqrt3 cos theta$. However, an important subtlety occurs when $theta = pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $pi/2 < theta < pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$frac12int_theta = pi/3^pi/2 ((sin theta)^2 - (sqrt3 cos theta)^2) , dtheta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$frac12 int_theta = pi/2^pi (sin theta)^2 , dtheta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.
answered Apr 1 at 4:14
heropupheropup
65.6k865104
$endgroup$
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 le theta le pi$.
You can see that when $0 le theta < pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $pi/3 le theta le pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = sin theta$ and outside the circle $r = sqrt3 cos theta$. However, an important subtlety occurs when $theta = pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $pi/2 < theta < pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$frac12int_theta = pi/3^pi/2 ((sin theta)^2 - (sqrt3 cos theta)^2) , dtheta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$frac12 int_theta = pi/2^pi (sin theta)^2 , dtheta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.
answered Apr 1 at 4:14
heropupheropup
65.6k865104
answered Apr 1 at 4:14
heropupheropup
65.6k865104
answered Apr 1 at 4:14
heropupheropup
65.6k865104
answered Apr 1 at 4:14
heropupheropup
65.6k865104
65.6k865104
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
add a comment |
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
$begingroup$
I had to keep coming back and showing this answer to a lot of people, but I finally understand. Thank you!
$endgroup$
– CodingMee
Apr 4 at 20:20
add a comment |
$begingroup$
The intersection for $x$ correspond to
$$sin(theta)cos(theta)=sqrt 3 cos(theta)cos(theta)$$
$$cos(theta)big(sin(theta)-sqrt 3 cos(theta)big)=0$$ then the two solutions corresponding to $cos(theta)=0$ and $sin(theta)-sqrt 3 cos(theta)=0$
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
The intersection for $x$ correspond to
$$sin(theta)cos(theta)=sqrt 3 cos(theta)cos(theta)$$
$$cos(theta)big(sin(theta)-sqrt 3 cos(theta)big)=0$$ then the two solutions corresponding to $cos(theta)=0$ and $sin(theta)-sqrt 3 cos(theta)=0$
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
The intersection for $x$ correspond to
$$sin(theta)cos(theta)=sqrt 3 cos(theta)cos(theta)$$
$$cos(theta)big(sin(theta)-sqrt 3 cos(theta)big)=0$$ then the two solutions corresponding to $cos(theta)=0$ and $sin(theta)-sqrt 3 cos(theta)=0$
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
The intersection for $x$ correspond to
$$sin(theta)cos(theta)=sqrt 3 cos(theta)cos(theta)$$
$$cos(theta)big(sin(theta)-sqrt 3 cos(theta)big)=0$$ then the two solutions corresponding to $cos(theta)=0$ and $sin(theta)-sqrt 3 cos(theta)=0$
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 1 at 4:25
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
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$begingroup$
The first integral finds the area of the smaller circle lying between the rays $theta=pi/3$ and $theta=pi$. But that is too much by the sector of the large circle between the rays $theta=pi/3$ and $theta=pi/2$, so that part is subtracted away by the second integral.
$endgroup$
– John Wayland Bales
Mar 31 at 23:37
$begingroup$
You're not telling me where the $pi/2$ is coming from...
$endgroup$
– CodingMee
Mar 31 at 23:58