Finding the surface area of $G=(x,y,z):x^2+y^2leq 1,0leq zleq x+1$

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Let $G=(x,y,z):x^2+y^2leq 1,0leq zleq x+1$ find the surface area of $partial G$




So it is a cylinder of radius $1$ bounded on the z-axis by $0$ and $x+1$. Can I say that because the maximum value of $x$ is $1$ then $zin[0,2]$



So we have $phi(u,r)=( cos(u), sin(u), r)$ so $|phi_utimes phi_r|=1$



So we have $int_0^2int_0^2pidudr=4pi$?




calculus multivariable-calculus surface-integrals




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  • You sure about the bounds for $r$?
    – Sorfosh
    3 hours ago














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Let $G=(x,y,z):x^2+y^2leq 1,0leq zleq x+1$ find the surface area of $partial G$




So it is a cylinder of radius $1$ bounded on the z-axis by $0$ and $x+1$. Can I say that because the maximum value of $x$ is $1$ then $zin[0,2]$



So we have $phi(u,r)=( cos(u), sin(u), r)$ so $|phi_utimes phi_r|=1$



So we have $int_0^2int_0^2pidudr=4pi$?




calculus multivariable-calculus surface-integrals




share|cite|improve this question





















  • You sure about the bounds for $r$?
    – Sorfosh
    3 hours ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






Let $G=(x,y,z):x^2+y^2leq 1,0leq zleq x+1$ find the surface area of $partial G$




So it is a cylinder of radius $1$ bounded on the z-axis by $0$ and $x+1$. Can I say that because the maximum value of $x$ is $1$ then $zin[0,2]$



So we have $phi(u,r)=( cos(u), sin(u), r)$ so $|phi_utimes phi_r|=1$



So we have $int_0^2int_0^2pidudr=4pi$?




calculus multivariable-calculus surface-integrals




share|cite|improve this question














Let $G=(x,y,z):x^2+y^2leq 1,0leq zleq x+1$ find the surface area of $partial G$




So it is a cylinder of radius $1$ bounded on the z-axis by $0$ and $x+1$. Can I say that because the maximum value of $x$ is $1$ then $zin[0,2]$



So we have $phi(u,r)=( cos(u), sin(u), r)$ so $|phi_utimes phi_r|=1$



So we have $int_0^2int_0^2pidudr=4pi$?





calculus multivariable-calculus surface-integrals





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share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Stefan4024

27.6k52874




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newhere

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  • You sure about the bounds for $r$?
    – Sorfosh
    3 hours ago
















  • You sure about the bounds for $r$?
    – Sorfosh
    3 hours ago















You sure about the bounds for $r$?
– Sorfosh
3 hours ago




You sure about the bounds for $r$?
– Sorfosh
3 hours ago










1 Answer
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First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.



Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) to (cos theta, sin theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:



$$int_0^2pi int_0^cos theta + 1dzdtheta = int_0^2pi (cos theta + 1) dtheta = sin theta + theta ; Bigg|_0^2pi = 2pi$$



Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $pi$. On the other side the top can be parametrized by $(x,y,z) to (r cos theta, r sin theta, r cos theta + 1)$. The magnitude of the normal vector would be $rsqrt2$. Finally the surface is given by



$$int_0^1int_0^2pirsqrt2dtheta dr = 2pi int_0^1 rsqrt2dr = sqrt2pi$$



Finally adding all the values gives us that the total surface would be $(3 + sqrt2)pi$






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    First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.



    Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) to (cos theta, sin theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:



    $$int_0^2pi int_0^cos theta + 1dzdtheta = int_0^2pi (cos theta + 1) dtheta = sin theta + theta ; Bigg|_0^2pi = 2pi$$



    Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $pi$. On the other side the top can be parametrized by $(x,y,z) to (r cos theta, r sin theta, r cos theta + 1)$. The magnitude of the normal vector would be $rsqrt2$. Finally the surface is given by



    $$int_0^1int_0^2pirsqrt2dtheta dr = 2pi int_0^1 rsqrt2dr = sqrt2pi$$



    Finally adding all the values gives us that the total surface would be $(3 + sqrt2)pi$






    share|cite|improve this answer



























      up vote
      2
      down vote













      First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.



      Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) to (cos theta, sin theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:



      $$int_0^2pi int_0^cos theta + 1dzdtheta = int_0^2pi (cos theta + 1) dtheta = sin theta + theta ; Bigg|_0^2pi = 2pi$$



      Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $pi$. On the other side the top can be parametrized by $(x,y,z) to (r cos theta, r sin theta, r cos theta + 1)$. The magnitude of the normal vector would be $rsqrt2$. Finally the surface is given by



      $$int_0^1int_0^2pirsqrt2dtheta dr = 2pi int_0^1 rsqrt2dr = sqrt2pi$$



      Finally adding all the values gives us that the total surface would be $(3 + sqrt2)pi$






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.



        Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) to (cos theta, sin theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:



        $$int_0^2pi int_0^cos theta + 1dzdtheta = int_0^2pi (cos theta + 1) dtheta = sin theta + theta ; Bigg|_0^2pi = 2pi$$



        Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $pi$. On the other side the top can be parametrized by $(x,y,z) to (r cos theta, r sin theta, r cos theta + 1)$. The magnitude of the normal vector would be $rsqrt2$. Finally the surface is given by



        $$int_0^1int_0^2pirsqrt2dtheta dr = 2pi int_0^1 rsqrt2dr = sqrt2pi$$



        Finally adding all the values gives us that the total surface would be $(3 + sqrt2)pi$






        share|cite|improve this answer















        First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.



        Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) to (cos theta, sin theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:



        $$int_0^2pi int_0^cos theta + 1dzdtheta = int_0^2pi (cos theta + 1) dtheta = sin theta + theta ; Bigg|_0^2pi = 2pi$$



        Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $pi$. On the other side the top can be parametrized by $(x,y,z) to (r cos theta, r sin theta, r cos theta + 1)$. The magnitude of the normal vector would be $rsqrt2$. Finally the surface is given by



        $$int_0^1int_0^2pirsqrt2dtheta dr = 2pi int_0^1 rsqrt2dr = sqrt2pi$$



        Finally adding all the values gives us that the total surface would be $(3 + sqrt2)pi$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








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        Stefan4024

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