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The gravitational potential experienced by a point on the $z$-axis due to a hemisphere


Identifying functions on the unit disk with functions on the upper hemisphereHow do I calculate the area of an ellipse projected onto a hemisphere?Triple Integration of gravitational potentialFind the potential at the center of the sphere.Trying to find the mass of sphere using density per unit areaUsing spherical coordinates find the limits of integration of the region inside a sphere with center $(a,0,0)$ and radius $a$Find flux through a sphereThe integral for gravitational potential of a uniform planetEvaluate the integral by changing to spherical coordinates as belowContinuity of potential function at an interior point













0












$begingroup$



A solid hemisphere of uniform density $rho$ occupies the region $$x^2+y^2+z^2le a^2,qquad zle0.$$ Find the gravitational potential due to the hemisphere at the point $(0,0,s)$ where $sgt0$. A uniform rod of density $m$ per unit length lies on the $z$-axis between $(0,0,c)$ and $(0,0,d)$ where $dgt cgt0$. Show that the force exerted on the rod by the hemisphere is $$psi(c)-psi(d)$$ where $$psi(lambda)=frac2pi Gmrho3left(fraca^3+lambda^3-left(a^2+lambda^2right)^3/2lambdaright).$$




So to determine the potential at $(0,0,s)$, I evaluated the following integral over the region $R$:



$$iiint_Rfracrho Gsqrtx^2+y^2+(z-s)^2,dx,dy,dz$$



Using a spherical substitution, the integral becomes:



$$int_r=0^aint_theta=fracpi2^piint_phi = 0^2 pifracr^2sin(theta)sqrtr^2+s^2-2rscos(theta)dphi dtheta dr$$



Which evaluates to
$$varphi(s)=fracpirho G3left(frac2a^3+3a^2s+s^3-sqrt(s^2+a^2)^3sright)$$



As for the second part, my understanding is that all we have to do is evaluate



$$mint_c^dvarphi^'(s),ds=mvarphi(d)-mvarphi(c)$$



My question is: is there anything wrong with my $varphi$ or is my method from the second half wrong?



Thanks!










share|cite|improve this question











$endgroup$
















    0












    $begingroup$



    A solid hemisphere of uniform density $rho$ occupies the region $$x^2+y^2+z^2le a^2,qquad zle0.$$ Find the gravitational potential due to the hemisphere at the point $(0,0,s)$ where $sgt0$. A uniform rod of density $m$ per unit length lies on the $z$-axis between $(0,0,c)$ and $(0,0,d)$ where $dgt cgt0$. Show that the force exerted on the rod by the hemisphere is $$psi(c)-psi(d)$$ where $$psi(lambda)=frac2pi Gmrho3left(fraca^3+lambda^3-left(a^2+lambda^2right)^3/2lambdaright).$$




    So to determine the potential at $(0,0,s)$, I evaluated the following integral over the region $R$:



    $$iiint_Rfracrho Gsqrtx^2+y^2+(z-s)^2,dx,dy,dz$$



    Using a spherical substitution, the integral becomes:



    $$int_r=0^aint_theta=fracpi2^piint_phi = 0^2 pifracr^2sin(theta)sqrtr^2+s^2-2rscos(theta)dphi dtheta dr$$



    Which evaluates to
    $$varphi(s)=fracpirho G3left(frac2a^3+3a^2s+s^3-sqrt(s^2+a^2)^3sright)$$



    As for the second part, my understanding is that all we have to do is evaluate



    $$mint_c^dvarphi^'(s),ds=mvarphi(d)-mvarphi(c)$$



    My question is: is there anything wrong with my $varphi$ or is my method from the second half wrong?



    Thanks!










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$



      A solid hemisphere of uniform density $rho$ occupies the region $$x^2+y^2+z^2le a^2,qquad zle0.$$ Find the gravitational potential due to the hemisphere at the point $(0,0,s)$ where $sgt0$. A uniform rod of density $m$ per unit length lies on the $z$-axis between $(0,0,c)$ and $(0,0,d)$ where $dgt cgt0$. Show that the force exerted on the rod by the hemisphere is $$psi(c)-psi(d)$$ where $$psi(lambda)=frac2pi Gmrho3left(fraca^3+lambda^3-left(a^2+lambda^2right)^3/2lambdaright).$$




      So to determine the potential at $(0,0,s)$, I evaluated the following integral over the region $R$:



      $$iiint_Rfracrho Gsqrtx^2+y^2+(z-s)^2,dx,dy,dz$$



      Using a spherical substitution, the integral becomes:



      $$int_r=0^aint_theta=fracpi2^piint_phi = 0^2 pifracr^2sin(theta)sqrtr^2+s^2-2rscos(theta)dphi dtheta dr$$



      Which evaluates to
      $$varphi(s)=fracpirho G3left(frac2a^3+3a^2s+s^3-sqrt(s^2+a^2)^3sright)$$



      As for the second part, my understanding is that all we have to do is evaluate



      $$mint_c^dvarphi^'(s),ds=mvarphi(d)-mvarphi(c)$$



      My question is: is there anything wrong with my $varphi$ or is my method from the second half wrong?



      Thanks!










      share|cite|improve this question











      $endgroup$





      A solid hemisphere of uniform density $rho$ occupies the region $$x^2+y^2+z^2le a^2,qquad zle0.$$ Find the gravitational potential due to the hemisphere at the point $(0,0,s)$ where $sgt0$. A uniform rod of density $m$ per unit length lies on the $z$-axis between $(0,0,c)$ and $(0,0,d)$ where $dgt cgt0$. Show that the force exerted on the rod by the hemisphere is $$psi(c)-psi(d)$$ where $$psi(lambda)=frac2pi Gmrho3left(fraca^3+lambda^3-left(a^2+lambda^2right)^3/2lambdaright).$$




      So to determine the potential at $(0,0,s)$, I evaluated the following integral over the region $R$:



      $$iiint_Rfracrho Gsqrtx^2+y^2+(z-s)^2,dx,dy,dz$$



      Using a spherical substitution, the integral becomes:



      $$int_r=0^aint_theta=fracpi2^piint_phi = 0^2 pifracr^2sin(theta)sqrtr^2+s^2-2rscos(theta)dphi dtheta dr$$



      Which evaluates to
      $$varphi(s)=fracpirho G3left(frac2a^3+3a^2s+s^3-sqrt(s^2+a^2)^3sright)$$



      As for the second part, my understanding is that all we have to do is evaluate



      $$mint_c^dvarphi^'(s),ds=mvarphi(d)-mvarphi(c)$$



      My question is: is there anything wrong with my $varphi$ or is my method from the second half wrong?



      Thanks!







      multivariable-calculus vector-fields






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 1:43









      Robert Howard

      2,2583935




      2,2583935










      asked Mar 16 at 23:49









      SulSul

      332114




      332114




















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