Verifying Green's theorem for a function












0














Let



$G = { (x,y) in mathbb{R}^2 : x^2+4y^2 >1, x^2+y^2 < 4 } $



$ int_G x^2+y^2 d(x,y) $



I want to verify Green's Theorem :
$ oint_{ partial G } f n ds = int_G operatorname{div}f, dx $



I solved for he righthand side:



$ int_G operatorname{div}f dx = frac{25}{32} pi $,



just by calculating the Integral for the circle and the ellipse separately and subtract.



For the lefthand side, I need to find $ f: G rightarrow mathbb{R^2} $ so that $ operatorname{div}f = x^2+y^2 $
that can be $ f=( frac{1}{3} x^3 , frac{1}{3} y^3 ) $



now my question is, how to proceed? :0
Do I have to split the region?
Is $ n= frac{1}{ sqrt{ (frac{1}{3} x^3 )^2+ ( frac{1}{3}y^3 )^2}} begin{pmatrix} frac{1}{3} x^3 \ frac{1}{3} y^3 end{pmatrix} $ ?



Looking forward to any help! :-)






real-analysis integration gaussian-integral greens-theorem







share|cite|improve this question
























  • Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
    – TurlocTheRed
    Nov 28 at 19:55










  • yes, it is $ div f = x^2+y^2 $
    – wondering1123
    Nov 28 at 20:04
















0














Let



$G = { (x,y) in mathbb{R}^2 : x^2+4y^2 >1, x^2+y^2 < 4 } $



$ int_G x^2+y^2 d(x,y) $



I want to verify Green's Theorem :
$ oint_{ partial G } f n ds = int_G operatorname{div}f, dx $



I solved for he righthand side:



$ int_G operatorname{div}f dx = frac{25}{32} pi $,



just by calculating the Integral for the circle and the ellipse separately and subtract.



For the lefthand side, I need to find $ f: G rightarrow mathbb{R^2} $ so that $ operatorname{div}f = x^2+y^2 $
that can be $ f=( frac{1}{3} x^3 , frac{1}{3} y^3 ) $



now my question is, how to proceed? :0
Do I have to split the region?
Is $ n= frac{1}{ sqrt{ (frac{1}{3} x^3 )^2+ ( frac{1}{3}y^3 )^2}} begin{pmatrix} frac{1}{3} x^3 \ frac{1}{3} y^3 end{pmatrix} $ ?



Looking forward to any help! :-)






real-analysis integration gaussian-integral greens-theorem







share|cite|improve this question
























  • Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
    – TurlocTheRed
    Nov 28 at 19:55










  • yes, it is $ div f = x^2+y^2 $
    – wondering1123
    Nov 28 at 20:04














0












0








0







Let



$G = { (x,y) in mathbb{R}^2 : x^2+4y^2 >1, x^2+y^2 < 4 } $



$ int_G x^2+y^2 d(x,y) $



I want to verify Green's Theorem :
$ oint_{ partial G } f n ds = int_G operatorname{div}f, dx $



I solved for he righthand side:



$ int_G operatorname{div}f dx = frac{25}{32} pi $,



just by calculating the Integral for the circle and the ellipse separately and subtract.



For the lefthand side, I need to find $ f: G rightarrow mathbb{R^2} $ so that $ operatorname{div}f = x^2+y^2 $
that can be $ f=( frac{1}{3} x^3 , frac{1}{3} y^3 ) $



now my question is, how to proceed? :0
Do I have to split the region?
Is $ n= frac{1}{ sqrt{ (frac{1}{3} x^3 )^2+ ( frac{1}{3}y^3 )^2}} begin{pmatrix} frac{1}{3} x^3 \ frac{1}{3} y^3 end{pmatrix} $ ?



Looking forward to any help! :-)






real-analysis integration gaussian-integral greens-theorem







share|cite|improve this question















Let



$G = { (x,y) in mathbb{R}^2 : x^2+4y^2 >1, x^2+y^2 < 4 } $



$ int_G x^2+y^2 d(x,y) $



I want to verify Green's Theorem :
$ oint_{ partial G } f n ds = int_G operatorname{div}f, dx $



I solved for he righthand side:



$ int_G operatorname{div}f dx = frac{25}{32} pi $,



just by calculating the Integral for the circle and the ellipse separately and subtract.



For the lefthand side, I need to find $ f: G rightarrow mathbb{R^2} $ so that $ operatorname{div}f = x^2+y^2 $
that can be $ f=( frac{1}{3} x^3 , frac{1}{3} y^3 ) $



now my question is, how to proceed? :0
Do I have to split the region?
Is $ n= frac{1}{ sqrt{ (frac{1}{3} x^3 )^2+ ( frac{1}{3}y^3 )^2}} begin{pmatrix} frac{1}{3} x^3 \ frac{1}{3} y^3 end{pmatrix} $ ?



Looking forward to any help! :-)






real-analysis integration gaussian-integral greens-theorem




real-analysis integration gaussian-integral greens-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 16:56









Bernard

117k637111




117k637111










asked Nov 28 at 16:47









wondering1123

10011




10011












  • Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
    – TurlocTheRed
    Nov 28 at 19:55










  • yes, it is $ div f = x^2+y^2 $
    – wondering1123
    Nov 28 at 20:04


















  • Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
    – TurlocTheRed
    Nov 28 at 19:55










  • yes, it is $ div f = x^2+y^2 $
    – wondering1123
    Nov 28 at 20:04
















Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
– TurlocTheRed
Nov 28 at 19:55




Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how?
– TurlocTheRed
Nov 28 at 19:55












yes, it is $ div f = x^2+y^2 $
– wondering1123
Nov 28 at 20:04




yes, it is $ div f = x^2+y^2 $
– wondering1123
Nov 28 at 20:04










2 Answers
2






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oldest

votes


















1














$mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $mathbf r(t) = (x(t), y(t))$, then $mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and
$$int_{partial D} mathbf F cdot frac {mathbf n} {|mathbf n|} ,ds =
int_a^b mathbf F cdot frac {mathbf n} {|mathbf n|} ,|mathbf r'| ,dt =
int_a^b mathbf F cdot mathbf n ,dt.$$
We have $mathbf F = (x^3/3, y^3/3), ,mathbf r_1(t) = (cos t, 1/2 sin t), ,mathbf r_2(t) = (2 cos t, 2 sin t)$, thus
$$begin{aligned}
&begin{aligned}
int_{partial D} mathbf F cdot mathbf n ,dt =
&int_0^{2 pi} left( frac {cos^3 t} 3, frac {sin^3 t} {24} right) cdot
left( -frac {cos t} 2, -sin t right) dt + {} \
&int_0^{2 pi} left( frac {8 cos^3 t} 3, frac {8 sin^3 t} {3} right) cdot
(2 cos t, 2 sin t) ,dt,
end{aligned} \
&int_D nabla cdot mathbf F ,dS =
int_0^{2 pi} int_{1/sqrt{cos^2 t + 4 sin^2 t}}^2 ,r^3 dr dt.
end{aligned}$$
Both are equal to $251 pi/32$.






share|cite|improve this answer





























    0














    $f=x^2+y^2=r^2=f=nabla cdot (r^3/4 hat{r})$



    So $vec{F}=frac{r^3}{4}hat{r}$



    The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.



    The curves here can be parameterized as



    $x=acos{theta}$



    $y=bsin{theta}$



    Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$



    The tangent line vector is $-asin{theta}hat{i}+bcos{theta}hat{j}$



    So the normal line vector is $bcos{theta}hat{i}+asin{theta}vec{j}$



    So the unit normal



    $hat{n}= frac{<bcos{theta}, asin{theta}>}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}}$



    $vec{r}=<acos{theta}, bsin{theta}>$



    So $vec{F}cdot hat{n}dA =frac{r^2}{4}frac{ab}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}} dtheta $



    You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2pi$






    share|cite|improve this answer























    • The norm of $n$ doesn't contain an $a b$ term.
      – Maxim
      Dec 2 at 9:14










    • Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
      – TurlocTheRed
      Dec 3 at 14:44










    • I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
      – Maxim
      Dec 3 at 16:07












    • $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
      – TurlocTheRed
      Dec 4 at 15:58












    • I asee what you mean about the norm. Corrected! Thanks!
      – TurlocTheRed
      Dec 4 at 16:00













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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    $mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $mathbf r(t) = (x(t), y(t))$, then $mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and
    $$int_{partial D} mathbf F cdot frac {mathbf n} {|mathbf n|} ,ds =
    int_a^b mathbf F cdot frac {mathbf n} {|mathbf n|} ,|mathbf r'| ,dt =
    int_a^b mathbf F cdot mathbf n ,dt.$$
    We have $mathbf F = (x^3/3, y^3/3), ,mathbf r_1(t) = (cos t, 1/2 sin t), ,mathbf r_2(t) = (2 cos t, 2 sin t)$, thus
    $$begin{aligned}
    &begin{aligned}
    int_{partial D} mathbf F cdot mathbf n ,dt =
    &int_0^{2 pi} left( frac {cos^3 t} 3, frac {sin^3 t} {24} right) cdot
    left( -frac {cos t} 2, -sin t right) dt + {} \
    &int_0^{2 pi} left( frac {8 cos^3 t} 3, frac {8 sin^3 t} {3} right) cdot
    (2 cos t, 2 sin t) ,dt,
    end{aligned} \
    &int_D nabla cdot mathbf F ,dS =
    int_0^{2 pi} int_{1/sqrt{cos^2 t + 4 sin^2 t}}^2 ,r^3 dr dt.
    end{aligned}$$
    Both are equal to $251 pi/32$.






    share|cite|improve this answer


























      1














      $mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $mathbf r(t) = (x(t), y(t))$, then $mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and
      $$int_{partial D} mathbf F cdot frac {mathbf n} {|mathbf n|} ,ds =
      int_a^b mathbf F cdot frac {mathbf n} {|mathbf n|} ,|mathbf r'| ,dt =
      int_a^b mathbf F cdot mathbf n ,dt.$$
      We have $mathbf F = (x^3/3, y^3/3), ,mathbf r_1(t) = (cos t, 1/2 sin t), ,mathbf r_2(t) = (2 cos t, 2 sin t)$, thus
      $$begin{aligned}
      &begin{aligned}
      int_{partial D} mathbf F cdot mathbf n ,dt =
      &int_0^{2 pi} left( frac {cos^3 t} 3, frac {sin^3 t} {24} right) cdot
      left( -frac {cos t} 2, -sin t right) dt + {} \
      &int_0^{2 pi} left( frac {8 cos^3 t} 3, frac {8 sin^3 t} {3} right) cdot
      (2 cos t, 2 sin t) ,dt,
      end{aligned} \
      &int_D nabla cdot mathbf F ,dS =
      int_0^{2 pi} int_{1/sqrt{cos^2 t + 4 sin^2 t}}^2 ,r^3 dr dt.
      end{aligned}$$
      Both are equal to $251 pi/32$.






      share|cite|improve this answer
























        1












        1








        1






        $mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $mathbf r(t) = (x(t), y(t))$, then $mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and
        $$int_{partial D} mathbf F cdot frac {mathbf n} {|mathbf n|} ,ds =
        int_a^b mathbf F cdot frac {mathbf n} {|mathbf n|} ,|mathbf r'| ,dt =
        int_a^b mathbf F cdot mathbf n ,dt.$$
        We have $mathbf F = (x^3/3, y^3/3), ,mathbf r_1(t) = (cos t, 1/2 sin t), ,mathbf r_2(t) = (2 cos t, 2 sin t)$, thus
        $$begin{aligned}
        &begin{aligned}
        int_{partial D} mathbf F cdot mathbf n ,dt =
        &int_0^{2 pi} left( frac {cos^3 t} 3, frac {sin^3 t} {24} right) cdot
        left( -frac {cos t} 2, -sin t right) dt + {} \
        &int_0^{2 pi} left( frac {8 cos^3 t} 3, frac {8 sin^3 t} {3} right) cdot
        (2 cos t, 2 sin t) ,dt,
        end{aligned} \
        &int_D nabla cdot mathbf F ,dS =
        int_0^{2 pi} int_{1/sqrt{cos^2 t + 4 sin^2 t}}^2 ,r^3 dr dt.
        end{aligned}$$
        Both are equal to $251 pi/32$.






        share|cite|improve this answer












        $mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $mathbf r(t) = (x(t), y(t))$, then $mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and
        $$int_{partial D} mathbf F cdot frac {mathbf n} {|mathbf n|} ,ds =
        int_a^b mathbf F cdot frac {mathbf n} {|mathbf n|} ,|mathbf r'| ,dt =
        int_a^b mathbf F cdot mathbf n ,dt.$$
        We have $mathbf F = (x^3/3, y^3/3), ,mathbf r_1(t) = (cos t, 1/2 sin t), ,mathbf r_2(t) = (2 cos t, 2 sin t)$, thus
        $$begin{aligned}
        &begin{aligned}
        int_{partial D} mathbf F cdot mathbf n ,dt =
        &int_0^{2 pi} left( frac {cos^3 t} 3, frac {sin^3 t} {24} right) cdot
        left( -frac {cos t} 2, -sin t right) dt + {} \
        &int_0^{2 pi} left( frac {8 cos^3 t} 3, frac {8 sin^3 t} {3} right) cdot
        (2 cos t, 2 sin t) ,dt,
        end{aligned} \
        &int_D nabla cdot mathbf F ,dS =
        int_0^{2 pi} int_{1/sqrt{cos^2 t + 4 sin^2 t}}^2 ,r^3 dr dt.
        end{aligned}$$
        Both are equal to $251 pi/32$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 at 9:09









        Maxim

        4,433219




        4,433219























            0














            $f=x^2+y^2=r^2=f=nabla cdot (r^3/4 hat{r})$



            So $vec{F}=frac{r^3}{4}hat{r}$



            The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.



            The curves here can be parameterized as



            $x=acos{theta}$



            $y=bsin{theta}$



            Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$



            The tangent line vector is $-asin{theta}hat{i}+bcos{theta}hat{j}$



            So the normal line vector is $bcos{theta}hat{i}+asin{theta}vec{j}$



            So the unit normal



            $hat{n}= frac{<bcos{theta}, asin{theta}>}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}}$



            $vec{r}=<acos{theta}, bsin{theta}>$



            So $vec{F}cdot hat{n}dA =frac{r^2}{4}frac{ab}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}} dtheta $



            You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2pi$






            share|cite|improve this answer























            • The norm of $n$ doesn't contain an $a b$ term.
              – Maxim
              Dec 2 at 9:14










            • Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
              – TurlocTheRed
              Dec 3 at 14:44










            • I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
              – Maxim
              Dec 3 at 16:07












            • $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
              – TurlocTheRed
              Dec 4 at 15:58












            • I asee what you mean about the norm. Corrected! Thanks!
              – TurlocTheRed
              Dec 4 at 16:00


















            0














            $f=x^2+y^2=r^2=f=nabla cdot (r^3/4 hat{r})$



            So $vec{F}=frac{r^3}{4}hat{r}$



            The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.



            The curves here can be parameterized as



            $x=acos{theta}$



            $y=bsin{theta}$



            Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$



            The tangent line vector is $-asin{theta}hat{i}+bcos{theta}hat{j}$



            So the normal line vector is $bcos{theta}hat{i}+asin{theta}vec{j}$



            So the unit normal



            $hat{n}= frac{<bcos{theta}, asin{theta}>}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}}$



            $vec{r}=<acos{theta}, bsin{theta}>$



            So $vec{F}cdot hat{n}dA =frac{r^2}{4}frac{ab}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}} dtheta $



            You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2pi$






            share|cite|improve this answer























            • The norm of $n$ doesn't contain an $a b$ term.
              – Maxim
              Dec 2 at 9:14










            • Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
              – TurlocTheRed
              Dec 3 at 14:44










            • I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
              – Maxim
              Dec 3 at 16:07












            • $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
              – TurlocTheRed
              Dec 4 at 15:58












            • I asee what you mean about the norm. Corrected! Thanks!
              – TurlocTheRed
              Dec 4 at 16:00
















            0












            0








            0






            $f=x^2+y^2=r^2=f=nabla cdot (r^3/4 hat{r})$



            So $vec{F}=frac{r^3}{4}hat{r}$



            The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.



            The curves here can be parameterized as



            $x=acos{theta}$



            $y=bsin{theta}$



            Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$



            The tangent line vector is $-asin{theta}hat{i}+bcos{theta}hat{j}$



            So the normal line vector is $bcos{theta}hat{i}+asin{theta}vec{j}$



            So the unit normal



            $hat{n}= frac{<bcos{theta}, asin{theta}>}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}}$



            $vec{r}=<acos{theta}, bsin{theta}>$



            So $vec{F}cdot hat{n}dA =frac{r^2}{4}frac{ab}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}} dtheta $



            You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2pi$






            share|cite|improve this answer














            $f=x^2+y^2=r^2=f=nabla cdot (r^3/4 hat{r})$



            So $vec{F}=frac{r^3}{4}hat{r}$



            The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.



            The curves here can be parameterized as



            $x=acos{theta}$



            $y=bsin{theta}$



            Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$



            The tangent line vector is $-asin{theta}hat{i}+bcos{theta}hat{j}$



            So the normal line vector is $bcos{theta}hat{i}+asin{theta}vec{j}$



            So the unit normal



            $hat{n}= frac{<bcos{theta}, asin{theta}>}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}}$



            $vec{r}=<acos{theta}, bsin{theta}>$



            So $vec{F}cdot hat{n}dA =frac{r^2}{4}frac{ab}{sqrt{b^2cos^2{theta}+a^2sin^2{theta}}} dtheta $



            You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2pi$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 at 16:15

























            answered Nov 28 at 20:12









            TurlocTheRed

            828311




            828311












            • The norm of $n$ doesn't contain an $a b$ term.
              – Maxim
              Dec 2 at 9:14










            • Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
              – TurlocTheRed
              Dec 3 at 14:44










            • I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
              – Maxim
              Dec 3 at 16:07












            • $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
              – TurlocTheRed
              Dec 4 at 15:58












            • I asee what you mean about the norm. Corrected! Thanks!
              – TurlocTheRed
              Dec 4 at 16:00




















            • The norm of $n$ doesn't contain an $a b$ term.
              – Maxim
              Dec 2 at 9:14










            • Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
              – TurlocTheRed
              Dec 3 at 14:44










            • I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
              – Maxim
              Dec 3 at 16:07












            • $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
              – TurlocTheRed
              Dec 4 at 15:58












            • I asee what you mean about the norm. Corrected! Thanks!
              – TurlocTheRed
              Dec 4 at 16:00


















            The norm of $n$ doesn't contain an $a b$ term.
            – Maxim
            Dec 2 at 9:14




            The norm of $n$ doesn't contain an $a b$ term.
            – Maxim
            Dec 2 at 9:14












            Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
            – TurlocTheRed
            Dec 3 at 14:44




            Correct. I took r from r^3 and multiplied it with $hat{r}$, then dotted that with $hat{n}$ producing the result.
            – TurlocTheRed
            Dec 3 at 14:44












            I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
            – Maxim
            Dec 3 at 16:07






            I'm saying that if $mathbf n = (b cos theta, a sin theta)$, then $|mathbf n| = sqrt{b^2 cos^2 theta + a^2 sin^2 theta}$. Also, the divergence of $r^3/3 ,hat {mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element?
            – Maxim
            Dec 3 at 16:07














            $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
            – TurlocTheRed
            Dec 4 at 15:58






            $nabla cdot vec{A}$ in cylindrical coordinates = $frac{partial A_r}{partial r}+frac{partial A_theta}{rpartial theta}+frac{partial A_z}{partial z}$ $vec{F}=r^3/3hat{r}$ So: $nabla cdot vec{F}=frac{partial(r^3/3)}{partial r}$ $nabla cdot vec{F}=r^2$
            – TurlocTheRed
            Dec 4 at 15:58














            I asee what you mean about the norm. Corrected! Thanks!
            – TurlocTheRed
            Dec 4 at 16:00






            I asee what you mean about the norm. Corrected! Thanks!
            – TurlocTheRed
            Dec 4 at 16:00




















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