Mixing
Double integral of off centre circle.
$begingroup$
I have the vector field $F = (3xy,-x)$ along the circle $c$ (counter clockwise) which has a radius $a$ and centre $(a,0)$.
I want to try and apply Green's Theorem to this, where I obtain $intint(-1 - 3x )dxdy$.
Now I try and use polar coordinates, where I replace $x = a + rcostheta$ and $dxdy =rdrdtheta$.
The limits of $r$ are from $0$ to $a$, and the limits of $theta$ from $0$ to $2pi$.
When I solve the integral I get $-a^{2}pi - 3a^{3}pi$ . Is this correct? I don't know the answer and im not entirely sure of my working. Many thanks!
calculus integration area greens-theorem
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
$endgroup$
add a comment |
$begingroup$
I have the vector field $F = (3xy,-x)$ along the circle $c$ (counter clockwise) which has a radius $a$ and centre $(a,0)$.
I want to try and apply Green's Theorem to this, where I obtain $intint(-1 - 3x )dxdy$.
Now I try and use polar coordinates, where I replace $x = a + rcostheta$ and $dxdy =rdrdtheta$.
The limits of $r$ are from $0$ to $a$, and the limits of $theta$ from $0$ to $2pi$.
When I solve the integral I get $-a^{2}pi - 3a^{3}pi$ . Is this correct? I don't know the answer and im not entirely sure of my working. Many thanks!
calculus integration area greens-theorem
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
$endgroup$
add a comment |
$begingroup$
I have the vector field $F = (3xy,-x)$ along the circle $c$ (counter clockwise) which has a radius $a$ and centre $(a,0)$.
I want to try and apply Green's Theorem to this, where I obtain $intint(-1 - 3x )dxdy$.
Now I try and use polar coordinates, where I replace $x = a + rcostheta$ and $dxdy =rdrdtheta$.
The limits of $r$ are from $0$ to $a$, and the limits of $theta$ from $0$ to $2pi$.
When I solve the integral I get $-a^{2}pi - 3a^{3}pi$ . Is this correct? I don't know the answer and im not entirely sure of my working. Many thanks!
calculus integration area greens-theorem
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
$endgroup$
I have the vector field $F = (3xy,-x)$ along the circle $c$ (counter clockwise) which has a radius $a$ and centre $(a,0)$.
I want to try and apply Green's Theorem to this, where I obtain $intint(-1 - 3x )dxdy$.
Now I try and use polar coordinates, where I replace $x = a + rcostheta$ and $dxdy =rdrdtheta$.
The limits of $r$ are from $0$ to $a$, and the limits of $theta$ from $0$ to $2pi$.
When I solve the integral I get $-a^{2}pi - 3a^{3}pi$ . Is this correct? I don't know the answer and im not entirely sure of my working. Many thanks!
calculus integration area greens-theorem
calculus integration area greens-theorem
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
edited Apr 21 '16 at 22:02
Arnaud D.
16.2k52444
16.2k52444
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
asked Apr 21 '16 at 21:51
stellarhawk 34stellarhawk 34
2726
2726
add a comment |
add a comment |
1 Answer
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$begingroup$
If you think about the integral as an average value, then
$$intint_Rf(x,y)d^2A=langle frangle_Rtimes A(R)$$
that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so
$$intint_R(-1-3x)d^2A=(-1-3a)cdotpi a^2=-pi a^2-3pi a^3$$
You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=rcostheta$, $y=rcostheta$, $0le rle2acostheta$ so that you might get
$$begin{align}int_{-frac{pi}2}^{frac{pi}2}int_0^{2acostheta}(-1-3rcostheta)r,dr,dtheta&=int_{-frac{pi}2}^{frac{pi}2}(-2a^2cos^2theta-8a^3cos^4theta)dtheta\
&=-2a^2left(frac{pi}2right)-8a^3left(frac{3pi}8right)=-pi a^2-3pi a^3end{align}$$
Where we have again used the average value of $frac12$ for $cos^2theta$ and of $frac38$ for $cos^4theta$.
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
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1 Answer
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1 Answer
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$begingroup$
If you think about the integral as an average value, then
$$intint_Rf(x,y)d^2A=langle frangle_Rtimes A(R)$$
that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so
$$intint_R(-1-3x)d^2A=(-1-3a)cdotpi a^2=-pi a^2-3pi a^3$$
You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=rcostheta$, $y=rcostheta$, $0le rle2acostheta$ so that you might get
$$begin{align}int_{-frac{pi}2}^{frac{pi}2}int_0^{2acostheta}(-1-3rcostheta)r,dr,dtheta&=int_{-frac{pi}2}^{frac{pi}2}(-2a^2cos^2theta-8a^3cos^4theta)dtheta\
&=-2a^2left(frac{pi}2right)-8a^3left(frac{3pi}8right)=-pi a^2-3pi a^3end{align}$$
Where we have again used the average value of $frac12$ for $cos^2theta$ and of $frac38$ for $cos^4theta$.
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
$endgroup$
add a comment |
$begingroup$
If you think about the integral as an average value, then
$$intint_Rf(x,y)d^2A=langle frangle_Rtimes A(R)$$
that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so
$$intint_R(-1-3x)d^2A=(-1-3a)cdotpi a^2=-pi a^2-3pi a^3$$
You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=rcostheta$, $y=rcostheta$, $0le rle2acostheta$ so that you might get
$$begin{align}int_{-frac{pi}2}^{frac{pi}2}int_0^{2acostheta}(-1-3rcostheta)r,dr,dtheta&=int_{-frac{pi}2}^{frac{pi}2}(-2a^2cos^2theta-8a^3cos^4theta)dtheta\
&=-2a^2left(frac{pi}2right)-8a^3left(frac{3pi}8right)=-pi a^2-3pi a^3end{align}$$
Where we have again used the average value of $frac12$ for $cos^2theta$ and of $frac38$ for $cos^4theta$.
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
$endgroup$
add a comment |
$begingroup$
If you think about the integral as an average value, then
$$intint_Rf(x,y)d^2A=langle frangle_Rtimes A(R)$$
that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so
$$intint_R(-1-3x)d^2A=(-1-3a)cdotpi a^2=-pi a^2-3pi a^3$$
You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=rcostheta$, $y=rcostheta$, $0le rle2acostheta$ so that you might get
$$begin{align}int_{-frac{pi}2}^{frac{pi}2}int_0^{2acostheta}(-1-3rcostheta)r,dr,dtheta&=int_{-frac{pi}2}^{frac{pi}2}(-2a^2cos^2theta-8a^3cos^4theta)dtheta\
&=-2a^2left(frac{pi}2right)-8a^3left(frac{3pi}8right)=-pi a^2-3pi a^3end{align}$$
Where we have again used the average value of $frac12$ for $cos^2theta$ and of $frac38$ for $cos^4theta$.
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
$endgroup$
If you think about the integral as an average value, then
$$intint_Rf(x,y)d^2A=langle frangle_Rtimes A(R)$$
that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so
$$intint_R(-1-3x)d^2A=(-1-3a)cdotpi a^2=-pi a^2-3pi a^3$$
You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=rcostheta$, $y=rcostheta$, $0le rle2acostheta$ so that you might get
$$begin{align}int_{-frac{pi}2}^{frac{pi}2}int_0^{2acostheta}(-1-3rcostheta)r,dr,dtheta&=int_{-frac{pi}2}^{frac{pi}2}(-2a^2cos^2theta-8a^3cos^4theta)dtheta\
&=-2a^2left(frac{pi}2right)-8a^3left(frac{3pi}8right)=-pi a^2-3pi a^3end{align}$$
Where we have again used the average value of $frac12$ for $cos^2theta$ and of $frac38$ for $cos^4theta$.
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
answered Apr 22 '16 at 3:42
user5713492user5713492
11.1k2919
11.1k2919
add a comment |
add a comment |
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