Volume after transformation.

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Consider $$ Omega: (x, y, z): x^2+y^2leq 1, 0leq zleq 2 $$ and the transform $$ T:(x, y, z)to(x, y+tan(alpha z), z) ,$$ where $ alphain (0, pi) $. What is the volume of $ T(Omega) $?




My attempt: I am trying to convert to the cylindrical coordinate system, but I can't find the region where I should integrate since it is transformed.






calculus integration multivariable-calculus







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    up vote
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    Consider $$ Omega: (x, y, z): x^2+y^2leq 1, 0leq zleq 2 $$ and the transform $$ T:(x, y, z)to(x, y+tan(alpha z), z) ,$$ where $ alphain (0, pi) $. What is the volume of $ T(Omega) $?




    My attempt: I am trying to convert to the cylindrical coordinate system, but I can't find the region where I should integrate since it is transformed.






    calculus integration multivariable-calculus







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1






      Consider $$ Omega: (x, y, z): x^2+y^2leq 1, 0leq zleq 2 $$ and the transform $$ T:(x, y, z)to(x, y+tan(alpha z), z) ,$$ where $ alphain (0, pi) $. What is the volume of $ T(Omega) $?




      My attempt: I am trying to convert to the cylindrical coordinate system, but I can't find the region where I should integrate since it is transformed.






      calculus integration multivariable-calculus







      share|cite|improve this question
















      Consider $$ Omega: (x, y, z): x^2+y^2leq 1, 0leq zleq 2 $$ and the transform $$ T:(x, y, z)to(x, y+tan(alpha z), z) ,$$ where $ alphain (0, pi) $. What is the volume of $ T(Omega) $?




      My attempt: I am trying to convert to the cylindrical coordinate system, but I can't find the region where I should integrate since it is transformed.






      calculus integration multivariable-calculus




      calculus integration multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 2 hours ago

























      asked 2 hours ago









      Yuchen

      874115




      874115




















          2 Answers
          2






          active

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          up vote
          2
          down vote



          accepted










          Fix $z_0 in [0,2]$, and look at the section $Omega_z_0 = Omega cap z = z_0$. This section is mapped to
          $$
          T(Omega_z_0) = (x,y,z_0) : x^2 + (y - tan(alpha z_0))^2 leq 1
          $$
          which is a unit circle in the plane $z = z_0$, centered at $(0, tan (alpha z_0))$. Hence, by the Cavalieri Principle, $T(Omega)$ has the same volume as $Omega$.






          share|cite|improve this answer






















          • Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
            – Yuchen
            1 hour ago






          • 1




            Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
            – Hugo
            1 hour ago










          • Yup, I see. Thanks again!
            – Yuchen
            1 hour ago

















          up vote
          4
          down vote













          Since the Jacobian for the transformation is



          $$|J|=beginvmatrix1&0&0\0&1&0\0&fracalphacosalpha z&1endvmatrix=1$$



          for $alphaneq fracpi2$ we have have that



          $$V_T(Omega)=V_Omega=2pi$$






          share|cite|improve this answer


















          • 1




            The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
            – orion
            1 hour ago










          • @orion Yes of course, I clarify that!
            – gimusi
            1 hour ago










          Your Answer




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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Fix $z_0 in [0,2]$, and look at the section $Omega_z_0 = Omega cap z = z_0$. This section is mapped to
          $$
          T(Omega_z_0) = (x,y,z_0) : x^2 + (y - tan(alpha z_0))^2 leq 1
          $$
          which is a unit circle in the plane $z = z_0$, centered at $(0, tan (alpha z_0))$. Hence, by the Cavalieri Principle, $T(Omega)$ has the same volume as $Omega$.






          share|cite|improve this answer






















          • Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
            – Yuchen
            1 hour ago






          • 1




            Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
            – Hugo
            1 hour ago










          • Yup, I see. Thanks again!
            – Yuchen
            1 hour ago














          up vote
          2
          down vote



          accepted










          Fix $z_0 in [0,2]$, and look at the section $Omega_z_0 = Omega cap z = z_0$. This section is mapped to
          $$
          T(Omega_z_0) = (x,y,z_0) : x^2 + (y - tan(alpha z_0))^2 leq 1
          $$
          which is a unit circle in the plane $z = z_0$, centered at $(0, tan (alpha z_0))$. Hence, by the Cavalieri Principle, $T(Omega)$ has the same volume as $Omega$.






          share|cite|improve this answer






















          • Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
            – Yuchen
            1 hour ago






          • 1




            Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
            – Hugo
            1 hour ago










          • Yup, I see. Thanks again!
            – Yuchen
            1 hour ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Fix $z_0 in [0,2]$, and look at the section $Omega_z_0 = Omega cap z = z_0$. This section is mapped to
          $$
          T(Omega_z_0) = (x,y,z_0) : x^2 + (y - tan(alpha z_0))^2 leq 1
          $$
          which is a unit circle in the plane $z = z_0$, centered at $(0, tan (alpha z_0))$. Hence, by the Cavalieri Principle, $T(Omega)$ has the same volume as $Omega$.






          share|cite|improve this answer














          Fix $z_0 in [0,2]$, and look at the section $Omega_z_0 = Omega cap z = z_0$. This section is mapped to
          $$
          T(Omega_z_0) = (x,y,z_0) : x^2 + (y - tan(alpha z_0))^2 leq 1
          $$
          which is a unit circle in the plane $z = z_0$, centered at $(0, tan (alpha z_0))$. Hence, by the Cavalieri Principle, $T(Omega)$ has the same volume as $Omega$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Hugo

          1064




          1064











          • Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
            – Yuchen
            1 hour ago






          • 1




            Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
            – Hugo
            1 hour ago










          • Yup, I see. Thanks again!
            – Yuchen
            1 hour ago
















          • Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
            – Yuchen
            1 hour ago






          • 1




            Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
            – Hugo
            1 hour ago










          • Yup, I see. Thanks again!
            – Yuchen
            1 hour ago















          Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
          – Yuchen
          1 hour ago




          Hi! Thank you for your nice answer! Could you revise the post? Since it is $ tan (alpha z_0) $ not $ z_0tanalpha $, I don't know why user 'amWhy' kept rejecting my attempt to revise your post.
          – Yuchen
          1 hour ago




          1




          1




          Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
          – Hugo
          1 hour ago




          Edited. The proof works in the same way, the discontinuty does not cause problems because a single plane has zero volume.
          – Hugo
          1 hour ago












          Yup, I see. Thanks again!
          – Yuchen
          1 hour ago




          Yup, I see. Thanks again!
          – Yuchen
          1 hour ago










          up vote
          4
          down vote













          Since the Jacobian for the transformation is



          $$|J|=beginvmatrix1&0&0\0&1&0\0&fracalphacosalpha z&1endvmatrix=1$$



          for $alphaneq fracpi2$ we have have that



          $$V_T(Omega)=V_Omega=2pi$$






          share|cite|improve this answer


















          • 1




            The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
            – orion
            1 hour ago










          • @orion Yes of course, I clarify that!
            – gimusi
            1 hour ago














          up vote
          4
          down vote













          Since the Jacobian for the transformation is



          $$|J|=beginvmatrix1&0&0\0&1&0\0&fracalphacosalpha z&1endvmatrix=1$$



          for $alphaneq fracpi2$ we have have that



          $$V_T(Omega)=V_Omega=2pi$$






          share|cite|improve this answer


















          • 1




            The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
            – orion
            1 hour ago










          • @orion Yes of course, I clarify that!
            – gimusi
            1 hour ago












          up vote
          4
          down vote










          up vote
          4
          down vote









          Since the Jacobian for the transformation is



          $$|J|=beginvmatrix1&0&0\0&1&0\0&fracalphacosalpha z&1endvmatrix=1$$



          for $alphaneq fracpi2$ we have have that



          $$V_T(Omega)=V_Omega=2pi$$






          share|cite|improve this answer














          Since the Jacobian for the transformation is



          $$|J|=beginvmatrix1&0&0\0&1&0\0&fracalphacosalpha z&1endvmatrix=1$$



          for $alphaneq fracpi2$ we have have that



          $$V_T(Omega)=V_Omega=2pi$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          gimusi

          75.4k73889




          75.4k73889







          • 1




            The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
            – orion
            1 hour ago










          • @orion Yes of course, I clarify that!
            – gimusi
            1 hour ago












          • 1




            The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
            – orion
            1 hour ago










          • @orion Yes of course, I clarify that!
            – gimusi
            1 hour ago







          1




          1




          The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
          – orion
          1 hour ago




          The Jacobian is 1 means that the volume is preserved (multiplied by 1) when you perform the transformation (it's just a fancy nonlinear shearing of slices perpendicular to z, in the y direction). So the volume can be calculated in the original coordinates, where it's just a cylinder. No integration needed.
          – orion
          1 hour ago












          @orion Yes of course, I clarify that!
          – gimusi
          1 hour ago




          @orion Yes of course, I clarify that!
          – gimusi
          1 hour ago

















           

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