How to find the surface area of the n-1 sphere?

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Q: Pick n>0 and suppose that Vol_n is the volume of the n-ball B(0,1), find the surface area of the n-1 sphere ∂B(ξ,ρ)?



I am totally stucked at where to start this problem. I have tried the transformation formula and the formula for finding surface areas itself. But neither of them seems to be related to the volume of the hypersphere. Need a hand! Thank you!



For more of you looking for answers for this problem, the link below is also useful..



https://www.quora.com/Geometry-Why-is-the-surface-area-of-a-sphere-the-derivative-of-the-volume-of-the-sphere




pde surface-integrals




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    up vote
    0
    down vote

    favorite












    Q: Pick n>0 and suppose that Vol_n is the volume of the n-ball B(0,1), find the surface area of the n-1 sphere ∂B(ξ,ρ)?



    I am totally stucked at where to start this problem. I have tried the transformation formula and the formula for finding surface areas itself. But neither of them seems to be related to the volume of the hypersphere. Need a hand! Thank you!



    For more of you looking for answers for this problem, the link below is also useful..



    https://www.quora.com/Geometry-Why-is-the-surface-area-of-a-sphere-the-derivative-of-the-volume-of-the-sphere




    pde surface-integrals




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Q: Pick n>0 and suppose that Vol_n is the volume of the n-ball B(0,1), find the surface area of the n-1 sphere ∂B(ξ,ρ)?



      I am totally stucked at where to start this problem. I have tried the transformation formula and the formula for finding surface areas itself. But neither of them seems to be related to the volume of the hypersphere. Need a hand! Thank you!



      For more of you looking for answers for this problem, the link below is also useful..



      https://www.quora.com/Geometry-Why-is-the-surface-area-of-a-sphere-the-derivative-of-the-volume-of-the-sphere




      pde surface-integrals




      share|cite|improve this question













      Q: Pick n>0 and suppose that Vol_n is the volume of the n-ball B(0,1), find the surface area of the n-1 sphere ∂B(ξ,ρ)?



      I am totally stucked at where to start this problem. I have tried the transformation formula and the formula for finding surface areas itself. But neither of them seems to be related to the volume of the hypersphere. Need a hand! Thank you!



      For more of you looking for answers for this problem, the link below is also useful..



      https://www.quora.com/Geometry-Why-is-the-surface-area-of-a-sphere-the-derivative-of-the-volume-of-the-sphere





      pde surface-integrals





      share|cite|improve this question












      share|cite|improve this question




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      edited Jul 26 at 8:16
























      asked Jul 26 at 5:46









      Evelyn Venne

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          1 Answer
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          down vote



          accepted










          Hint: Increasing $rho$ by a little bit, results in a change in volume which is proportional to the surface area (this is another way of saying that the derivative of the volume is the surface area)



          Check that when $n=2,3$ , the familiar formulas do obey this rule: $frac ddr(pi r^2)=2pi r$ and $frac ddr...
          (frac43pi r^3)=4pi r^2$.



          So, we get




          $partial B(xi,rho)=frac ddrho (operatorname Vol_n rho^n)=ncdot operatorname Vol_ncdot rho^n-1$







          share|cite|improve this answer





















          • Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
            – Evelyn Venne
            Jul 26 at 7:10










          • Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
            – Chris Custer
            Jul 26 at 7:19










          • For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
            – Chris Custer
            Jul 26 at 7:31










          • Cool that makes sense thanks heaps!
            – Evelyn Venne
            Jul 26 at 8:14










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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Hint: Increasing $rho$ by a little bit, results in a change in volume which is proportional to the surface area (this is another way of saying that the derivative of the volume is the surface area)



          Check that when $n=2,3$ , the familiar formulas do obey this rule: $frac ddr(pi r^2)=2pi r$ and $frac ddr...
          (frac43pi r^3)=4pi r^2$.



          So, we get




          $partial B(xi,rho)=frac ddrho (operatorname Vol_n rho^n)=ncdot operatorname Vol_ncdot rho^n-1$







          share|cite|improve this answer





















          • Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
            – Evelyn Venne
            Jul 26 at 7:10










          • Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
            – Chris Custer
            Jul 26 at 7:19










          • For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
            – Chris Custer
            Jul 26 at 7:31










          • Cool that makes sense thanks heaps!
            – Evelyn Venne
            Jul 26 at 8:14














          up vote
          1
          down vote



          accepted










          Hint: Increasing $rho$ by a little bit, results in a change in volume which is proportional to the surface area (this is another way of saying that the derivative of the volume is the surface area)



          Check that when $n=2,3$ , the familiar formulas do obey this rule: $frac ddr(pi r^2)=2pi r$ and $frac ddr...
          (frac43pi r^3)=4pi r^2$.



          So, we get




          $partial B(xi,rho)=frac ddrho (operatorname Vol_n rho^n)=ncdot operatorname Vol_ncdot rho^n-1$







          share|cite|improve this answer





















          • Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
            – Evelyn Venne
            Jul 26 at 7:10










          • Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
            – Chris Custer
            Jul 26 at 7:19










          • For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
            – Chris Custer
            Jul 26 at 7:31










          • Cool that makes sense thanks heaps!
            – Evelyn Venne
            Jul 26 at 8:14












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint: Increasing $rho$ by a little bit, results in a change in volume which is proportional to the surface area (this is another way of saying that the derivative of the volume is the surface area)



          Check that when $n=2,3$ , the familiar formulas do obey this rule: $frac ddr(pi r^2)=2pi r$ and $frac ddr...
          (frac43pi r^3)=4pi r^2$.



          So, we get




          $partial B(xi,rho)=frac ddrho (operatorname Vol_n rho^n)=ncdot operatorname Vol_ncdot rho^n-1$







          share|cite|improve this answer













          Hint: Increasing $rho$ by a little bit, results in a change in volume which is proportional to the surface area (this is another way of saying that the derivative of the volume is the surface area)



          Check that when $n=2,3$ , the familiar formulas do obey this rule: $frac ddr(pi r^2)=2pi r$ and $frac ddr...
          (frac43pi r^3)=4pi r^2$.



          So, we get




          $partial B(xi,rho)=frac ddrho (operatorname Vol_n rho^n)=ncdot operatorname Vol_ncdot rho^n-1$








          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 6:30









          Chris Custer

          5,2982622




          5,2982622











          • Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
            – Evelyn Venne
            Jul 26 at 7:10










          • Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
            – Chris Custer
            Jul 26 at 7:19










          • For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
            – Chris Custer
            Jul 26 at 7:31










          • Cool that makes sense thanks heaps!
            – Evelyn Venne
            Jul 26 at 8:14
















          • Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
            – Evelyn Venne
            Jul 26 at 7:10










          • Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
            – Chris Custer
            Jul 26 at 7:19










          • For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
            – Chris Custer
            Jul 26 at 7:31










          • Cool that makes sense thanks heaps!
            – Evelyn Venne
            Jul 26 at 8:14















          Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
          – Evelyn Venne
          Jul 26 at 7:10




          Thanks for your answer. But how could you see that the change in volume is proportional to the surface area, and how could you realise that it is the derivative relationship?..
          – Evelyn Venne
          Jul 26 at 7:10












          Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
          – Chris Custer
          Jul 26 at 7:19




          Good question. It's actually only true for "infinitesimal" changes... Here's a link: askamathematician.com/2013/02/…
          – Chris Custer
          Jul 26 at 7:19












          For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
          – Chris Custer
          Jul 26 at 7:31




          For my part, it's evidently true in lower dimensions... and this behavior generalizes... check out the link to higher dimensions in the link above.
          – Chris Custer
          Jul 26 at 7:31












          Cool that makes sense thanks heaps!
          – Evelyn Venne
          Jul 26 at 8:14




          Cool that makes sense thanks heaps!
          – Evelyn Venne
          Jul 26 at 8:14












           

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