surface area with integrals












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$begingroup$


I'm working on a problem in my textbook and am confused on how to set up the integral.



"Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.



I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.






integration multivariable-calculus surface-integrals







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm working on a problem in my textbook and am confused on how to set up the integral.



    "Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.



    I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.






    integration multivariable-calculus surface-integrals







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm working on a problem in my textbook and am confused on how to set up the integral.



      "Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.



      I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.






      integration multivariable-calculus surface-integrals







      share|cite|improve this question











      $endgroup$




      I'm working on a problem in my textbook and am confused on how to set up the integral.



      "Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.



      I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.






      integration multivariable-calculus surface-integrals




      integration multivariable-calculus surface-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 21:47









      user376343

      3,4233826




      3,4233826










      asked Jan 8 at 21:43









      marbeiiimarbeiii

      113




      113






















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          $begingroup$

          You need to parameterize the integration region. One possible choice is to take
          $$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
          with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
          $$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$



          Then the surface element is
          $$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
          The area of the region of interest (call it $R$) is given by the surface integral,
          $$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$






          share|cite|improve this answer









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            $begingroup$

            You need to parameterize the integration region. One possible choice is to take
            $$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
            with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
            $$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$



            Then the surface element is
            $$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
            The area of the region of interest (call it $R$) is given by the surface integral,
            $$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$






            share|cite|improve this answer









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              0












              $begingroup$

              You need to parameterize the integration region. One possible choice is to take
              $$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
              with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
              $$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$



              Then the surface element is
              $$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
              The area of the region of interest (call it $R$) is given by the surface integral,
              $$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You need to parameterize the integration region. One possible choice is to take
                $$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
                with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
                $$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$



                Then the surface element is
                $$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
                The area of the region of interest (call it $R$) is given by the surface integral,
                $$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$






                share|cite|improve this answer









                $endgroup$



                You need to parameterize the integration region. One possible choice is to take
                $$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
                with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
                $$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$



                Then the surface element is
                $$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
                The area of the region of interest (call it $R$) is given by the surface integral,
                $$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 21:58









                user170231user170231

                4,04711429




                4,04711429






























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