Mixing
Cardioid and Integration
$begingroup$
How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?
I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.
Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?
multivariable-calculus
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
$endgroup$
add a comment |
$begingroup$
How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?
I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.
Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?
multivariable-calculus
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
$endgroup$
$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59
add a comment |
$begingroup$
How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?
I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.
Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?
multivariable-calculus
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
$endgroup$
How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?
I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.
Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?
multivariable-calculus
multivariable-calculus
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
asked Jan 30 at 21:50
Shashank DwivediShashank Dwivedi
735
735
$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59
add a comment |
$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59
$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59
$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59
add a comment |
1 Answer
1
active
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votes
$begingroup$
Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.
Anyway,
convert to polar
$x = rcos theta\
y = rsintheta$
limits for theta.
$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$
$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$
If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.
$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.
Anyway,
convert to polar
$x = rcos theta\
y = rsintheta$
limits for theta.
$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$
$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$
If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.
$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.
Anyway,
convert to polar
$x = rcos theta\
y = rsintheta$
limits for theta.
$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$
$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$
If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.
$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.
Anyway,
convert to polar
$x = rcos theta\
y = rsintheta$
limits for theta.
$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$
$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$
If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.
$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
$endgroup$
Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.
Anyway,
convert to polar
$x = rcos theta\
y = rsintheta$
limits for theta.
$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$
$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$
If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.
$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
answered Jan 30 at 22:14
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59