Find the volume of the solid generated by revolving the region bounded by $y=sqrt{9-x^2}$ and $y=0$ about the...












0












$begingroup$


The answer for this question is apparently $36pi$, but I cannot get that answer. I keep getting $18pi$, and I have a feeling my professor made a mistake on this. Can I get some clarification?






integration volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
    $endgroup$
    – Doug M
    Dec 12 '18 at 2:40












  • $begingroup$
    When you do [-3,3] you get 0, the terms cancel eachother out.
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:09










  • $begingroup$
    Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:25










  • $begingroup$
    Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:46
















0












$begingroup$


The answer for this question is apparently $36pi$, but I cannot get that answer. I keep getting $18pi$, and I have a feeling my professor made a mistake on this. Can I get some clarification?






integration volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
    $endgroup$
    – Doug M
    Dec 12 '18 at 2:40












  • $begingroup$
    When you do [-3,3] you get 0, the terms cancel eachother out.
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:09










  • $begingroup$
    Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:25










  • $begingroup$
    Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:46














0












0








0





$begingroup$


The answer for this question is apparently $36pi$, but I cannot get that answer. I keep getting $18pi$, and I have a feeling my professor made a mistake on this. Can I get some clarification?






integration volume







share|cite|improve this question











$endgroup$




The answer for this question is apparently $36pi$, but I cannot get that answer. I keep getting $18pi$, and I have a feeling my professor made a mistake on this. Can I get some clarification?






integration volume




integration volume






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 3:37









bjcolby15

1,29711016




1,29711016










asked Dec 12 '18 at 2:36









Luke DLuke D

796




796












  • $begingroup$
    What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
    $endgroup$
    – Doug M
    Dec 12 '18 at 2:40












  • $begingroup$
    When you do [-3,3] you get 0, the terms cancel eachother out.
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:09










  • $begingroup$
    Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:25










  • $begingroup$
    Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:46


















  • $begingroup$
    What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
    $endgroup$
    – Doug M
    Dec 12 '18 at 2:40












  • $begingroup$
    When you do [-3,3] you get 0, the terms cancel eachother out.
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:09










  • $begingroup$
    Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:25










  • $begingroup$
    Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
    $endgroup$
    – Luke D
    Dec 12 '18 at 3:46
















$begingroup$
What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
$endgroup$
– Doug M
Dec 12 '18 at 2:40






$begingroup$
What are your limits of integration $[0, 3]$ or $[-3,3]$? That could well be the difference. You should be in the habit of showing your work, it helps us to give better answers.
$endgroup$
– Doug M
Dec 12 '18 at 2:40














$begingroup$
When you do [-3,3] you get 0, the terms cancel eachother out.
$endgroup$
– Luke D
Dec 12 '18 at 3:09




$begingroup$
When you do [-3,3] you get 0, the terms cancel eachother out.
$endgroup$
– Luke D
Dec 12 '18 at 3:09












$begingroup$
Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:25




$begingroup$
Well, the volume of a sphere with radius $R$ is $frac{4pi}{3}R^3$, so if $R=3$ the volume is $36pi$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:25












$begingroup$
Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
$endgroup$
– Luke D
Dec 12 '18 at 3:46




$begingroup$
Ah ok, what I did was I took the integral from [0,3] then solved. After factoring and taking the anti derivative from y=sqrt(y-x^2) , I got 9x - 1/3 *x^3 which gave me 18pi. I thought the radius is found by using the originial equation given?
$endgroup$
– Luke D
Dec 12 '18 at 3:46










2 Answers
2






active

oldest

votes


















2












$begingroup$

You may calculate the volume as follows:



$$pi int_{-3}^3 y^2 dx = pi int_{-3}^3 9 - x^2 ; dx = pi left[ 9x - frac{x^3}{3}right]_{-3}^3 = pi left[ 27 - 9 +27 - 9right] = 36 pi$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can also use symmetry to simplify the integral, i.e. when you have



    $$int_{-a}^a f(x) dx$$ you can rewrite it as $$2 int_{0}^a f(x) dx.$$



    Note: randomgirl is correct. I had mistakenly used the shell method (which is $pi int_a^{b} x f(x) dx$) rather than the disk method (which is $pi int_a^{b} f(x)^2 dx$). Sorry about that!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Think you put an extra x in it.
      $endgroup$
      – randomgirl
      Dec 12 '18 at 3:03










    • $begingroup$
      Oops and squaring the radius
      $endgroup$
      – randomgirl
      Dec 12 '18 at 3:03










    • $begingroup$
      Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
      $endgroup$
      – bjcolby15
      Dec 14 '18 at 0:12











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You may calculate the volume as follows:



    $$pi int_{-3}^3 y^2 dx = pi int_{-3}^3 9 - x^2 ; dx = pi left[ 9x - frac{x^3}{3}right]_{-3}^3 = pi left[ 27 - 9 +27 - 9right] = 36 pi$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You may calculate the volume as follows:



      $$pi int_{-3}^3 y^2 dx = pi int_{-3}^3 9 - x^2 ; dx = pi left[ 9x - frac{x^3}{3}right]_{-3}^3 = pi left[ 27 - 9 +27 - 9right] = 36 pi$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You may calculate the volume as follows:



        $$pi int_{-3}^3 y^2 dx = pi int_{-3}^3 9 - x^2 ; dx = pi left[ 9x - frac{x^3}{3}right]_{-3}^3 = pi left[ 27 - 9 +27 - 9right] = 36 pi$$






        share|cite|improve this answer









        $endgroup$



        You may calculate the volume as follows:



        $$pi int_{-3}^3 y^2 dx = pi int_{-3}^3 9 - x^2 ; dx = pi left[ 9x - frac{x^3}{3}right]_{-3}^3 = pi left[ 27 - 9 +27 - 9right] = 36 pi$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 3:53









        trancelocationtrancelocation

        10.8k1723




        10.8k1723























            0












            $begingroup$

            You can also use symmetry to simplify the integral, i.e. when you have



            $$int_{-a}^a f(x) dx$$ you can rewrite it as $$2 int_{0}^a f(x) dx.$$



            Note: randomgirl is correct. I had mistakenly used the shell method (which is $pi int_a^{b} x f(x) dx$) rather than the disk method (which is $pi int_a^{b} f(x)^2 dx$). Sorry about that!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Think you put an extra x in it.
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Oops and squaring the radius
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
              $endgroup$
              – bjcolby15
              Dec 14 '18 at 0:12
















            0












            $begingroup$

            You can also use symmetry to simplify the integral, i.e. when you have



            $$int_{-a}^a f(x) dx$$ you can rewrite it as $$2 int_{0}^a f(x) dx.$$



            Note: randomgirl is correct. I had mistakenly used the shell method (which is $pi int_a^{b} x f(x) dx$) rather than the disk method (which is $pi int_a^{b} f(x)^2 dx$). Sorry about that!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Think you put an extra x in it.
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Oops and squaring the radius
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
              $endgroup$
              – bjcolby15
              Dec 14 '18 at 0:12














            0












            0








            0





            $begingroup$

            You can also use symmetry to simplify the integral, i.e. when you have



            $$int_{-a}^a f(x) dx$$ you can rewrite it as $$2 int_{0}^a f(x) dx.$$



            Note: randomgirl is correct. I had mistakenly used the shell method (which is $pi int_a^{b} x f(x) dx$) rather than the disk method (which is $pi int_a^{b} f(x)^2 dx$). Sorry about that!






            share|cite|improve this answer











            $endgroup$



            You can also use symmetry to simplify the integral, i.e. when you have



            $$int_{-a}^a f(x) dx$$ you can rewrite it as $$2 int_{0}^a f(x) dx.$$



            Note: randomgirl is correct. I had mistakenly used the shell method (which is $pi int_a^{b} x f(x) dx$) rather than the disk method (which is $pi int_a^{b} f(x)^2 dx$). Sorry about that!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 14 '18 at 0:10

























            answered Dec 12 '18 at 3:00









            bjcolby15bjcolby15

            1,29711016




            1,29711016












            • $begingroup$
              Think you put an extra x in it.
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Oops and squaring the radius
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
              $endgroup$
              – bjcolby15
              Dec 14 '18 at 0:12


















            • $begingroup$
              Think you put an extra x in it.
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Oops and squaring the radius
              $endgroup$
              – randomgirl
              Dec 12 '18 at 3:03










            • $begingroup$
              Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
              $endgroup$
              – bjcolby15
              Dec 14 '18 at 0:12
















            $begingroup$
            Think you put an extra x in it.
            $endgroup$
            – randomgirl
            Dec 12 '18 at 3:03




            $begingroup$
            Think you put an extra x in it.
            $endgroup$
            – randomgirl
            Dec 12 '18 at 3:03












            $begingroup$
            Oops and squaring the radius
            $endgroup$
            – randomgirl
            Dec 12 '18 at 3:03




            $begingroup$
            Oops and squaring the radius
            $endgroup$
            – randomgirl
            Dec 12 '18 at 3:03












            $begingroup$
            Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
            $endgroup$
            – bjcolby15
            Dec 14 '18 at 0:12




            $begingroup$
            Thanks, randomgirl! I got the two revolution methods mixed up - the disk method is the proper one.
            $endgroup$
            – bjcolby15
            Dec 14 '18 at 0:12


















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