Show that the solid generated by the revolution of the region has finite volume but infinite surface area

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Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.






calculus solid-of-revolution







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  • 1




    Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
    – Jyrki Lahtonen
    3 hours ago






  • 1




    See also here
    – Jyrki Lahtonen
    2 hours ago










  • thanks @Jyrki Lahtonen. Yeah i did type mistake :)
    – md emon
    2 hours ago














up vote
3
down vote

favorite












Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.






calculus solid-of-revolution







share|cite|improve this question



















  • 1




    Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
    – Jyrki Lahtonen
    3 hours ago






  • 1




    See also here
    – Jyrki Lahtonen
    2 hours ago










  • thanks @Jyrki Lahtonen. Yeah i did type mistake :)
    – md emon
    2 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.






calculus solid-of-revolution







share|cite|improve this question















Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.






calculus solid-of-revolution




calculus solid-of-revolution






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share|cite|improve this question













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edited 2 hours ago

























asked 3 hours ago









md emon

17111




17111







  • 1




    Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
    – Jyrki Lahtonen
    3 hours ago






  • 1




    See also here
    – Jyrki Lahtonen
    2 hours ago










  • thanks @Jyrki Lahtonen. Yeah i did type mistake :)
    – md emon
    2 hours ago












  • 1




    Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
    – Jyrki Lahtonen
    3 hours ago






  • 1




    See also here
    – Jyrki Lahtonen
    2 hours ago










  • thanks @Jyrki Lahtonen. Yeah i did type mistake :)
    – md emon
    2 hours ago







1




1




Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago




Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago




1




1




See also here
– Jyrki Lahtonen
2 hours ago




See also here
– Jyrki Lahtonen
2 hours ago












thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago




thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago










2 Answers
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up vote
3
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For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$






share|cite|improve this answer




















  • Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
    – md emon
    2 hours ago










  • Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
    – Andrei
    2 hours ago


















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2
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Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.






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    2 Answers
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    2 Answers
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    active

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    up vote
    3
    down vote













    For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$






    share|cite|improve this answer




















    • Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
      – md emon
      2 hours ago










    • Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
      – Andrei
      2 hours ago















    up vote
    3
    down vote













    For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$






    share|cite|improve this answer




















    • Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
      – md emon
      2 hours ago










    • Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
      – Andrei
      2 hours ago













    up vote
    3
    down vote










    up vote
    3
    down vote









    For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$






    share|cite|improve this answer












    For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    Andrei

    8,4632923




    8,4632923











    • Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
      – md emon
      2 hours ago










    • Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
      – Andrei
      2 hours ago

















    • Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
      – md emon
      2 hours ago










    • Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
      – Andrei
      2 hours ago
















    Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
    – md emon
    2 hours ago




    Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
    – md emon
    2 hours ago












    Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
    – Andrei
    2 hours ago





    Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
    – Andrei
    2 hours ago











    up vote
    2
    down vote













    Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
    $$int_1^inftyfrac 1x,dx$$
    which diverges.






    share|cite|improve this answer
























      up vote
      2
      down vote













      Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
      $$int_1^inftyfrac 1x,dx$$
      which diverges.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
        $$int_1^inftyfrac 1x,dx$$
        which diverges.






        share|cite|improve this answer












        Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
        $$int_1^inftyfrac 1x,dx$$
        which diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Parcly Taxel

        34k136890




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