Volume of Revolution for $y=1-x^2$ and $y=2x$












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There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?






calculus solid-of-revolution







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    $begingroup$


    There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?






    calculus solid-of-revolution







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?






      calculus solid-of-revolution







      share|cite|improve this question











      $endgroup$




      There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?






      calculus solid-of-revolution




      calculus solid-of-revolution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 20:14







      user3896106

















      asked Dec 8 '18 at 20:09









      user3896106user3896106

      83




      83






















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
            $endgroup$
            – user3896106
            Dec 8 '18 at 20:20












          • $begingroup$
            No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
            $endgroup$
            – Chris Custer
            Dec 8 '18 at 20:43



















          -1












          $begingroup$

          The integral



          $$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$



          represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.



          Refer also to the following sketch



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you decide of the axis of rotation from the OP?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 21:57










          • $begingroup$
            @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:01










          • $begingroup$
            So you use the given answer to decide what the missing information in the question actually is?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:17










          • $begingroup$
            @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:22










          • $begingroup$
            Yes, I agree...
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:44











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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
            $endgroup$
            – user3896106
            Dec 8 '18 at 20:20












          • $begingroup$
            No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
            $endgroup$
            – Chris Custer
            Dec 8 '18 at 20:43
















          0












          $begingroup$

          Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
            $endgroup$
            – user3896106
            Dec 8 '18 at 20:20












          • $begingroup$
            No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
            $endgroup$
            – Chris Custer
            Dec 8 '18 at 20:43














          0












          0








          0





          $begingroup$

          Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.






          share|cite|improve this answer









          $endgroup$



          Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 20:16









          Chris CusterChris Custer

          14.2k3827




          14.2k3827












          • $begingroup$
            It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
            $endgroup$
            – user3896106
            Dec 8 '18 at 20:20












          • $begingroup$
            No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
            $endgroup$
            – Chris Custer
            Dec 8 '18 at 20:43


















          • $begingroup$
            It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
            $endgroup$
            – user3896106
            Dec 8 '18 at 20:20












          • $begingroup$
            No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
            $endgroup$
            – Chris Custer
            Dec 8 '18 at 20:43
















          $begingroup$
          It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
          $endgroup$
          – user3896106
          Dec 8 '18 at 20:20






          $begingroup$
          It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
          $endgroup$
          – user3896106
          Dec 8 '18 at 20:20














          $begingroup$
          No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
          $endgroup$
          – Chris Custer
          Dec 8 '18 at 20:43




          $begingroup$
          No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
          $endgroup$
          – Chris Custer
          Dec 8 '18 at 20:43











          -1












          $begingroup$

          The integral



          $$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$



          represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.



          Refer also to the following sketch



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you decide of the axis of rotation from the OP?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 21:57










          • $begingroup$
            @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:01










          • $begingroup$
            So you use the given answer to decide what the missing information in the question actually is?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:17










          • $begingroup$
            @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:22










          • $begingroup$
            Yes, I agree...
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:44
















          -1












          $begingroup$

          The integral



          $$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$



          represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.



          Refer also to the following sketch



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you decide of the axis of rotation from the OP?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 21:57










          • $begingroup$
            @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:01










          • $begingroup$
            So you use the given answer to decide what the missing information in the question actually is?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:17










          • $begingroup$
            @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:22










          • $begingroup$
            Yes, I agree...
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:44














          -1












          -1








          -1





          $begingroup$

          The integral



          $$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$



          represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.



          Refer also to the following sketch



          enter image description here






          share|cite|improve this answer











          $endgroup$



          The integral



          $$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$



          represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.



          Refer also to the following sketch



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 20:20

























          answered Dec 8 '18 at 20:13









          gimusigimusi

          93k84594




          93k84594












          • $begingroup$
            How do you decide of the axis of rotation from the OP?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 21:57










          • $begingroup$
            @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:01










          • $begingroup$
            So you use the given answer to decide what the missing information in the question actually is?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:17










          • $begingroup$
            @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:22










          • $begingroup$
            Yes, I agree...
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:44


















          • $begingroup$
            How do you decide of the axis of rotation from the OP?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 21:57










          • $begingroup$
            @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:01










          • $begingroup$
            So you use the given answer to decide what the missing information in the question actually is?
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:17










          • $begingroup$
            @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
            $endgroup$
            – gimusi
            Dec 8 '18 at 22:22










          • $begingroup$
            Yes, I agree...
            $endgroup$
            – DJohnM
            Dec 8 '18 at 22:44
















          $begingroup$
          How do you decide of the axis of rotation from the OP?
          $endgroup$
          – DJohnM
          Dec 8 '18 at 21:57




          $begingroup$
          How do you decide of the axis of rotation from the OP?
          $endgroup$
          – DJohnM
          Dec 8 '18 at 21:57












          $begingroup$
          @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
          $endgroup$
          – gimusi
          Dec 8 '18 at 22:01




          $begingroup$
          @DJohnM For the given set up otherwise the integral would be different and in 2 parts.
          $endgroup$
          – gimusi
          Dec 8 '18 at 22:01












          $begingroup$
          So you use the given answer to decide what the missing information in the question actually is?
          $endgroup$
          – DJohnM
          Dec 8 '18 at 22:17




          $begingroup$
          So you use the given answer to decide what the missing information in the question actually is?
          $endgroup$
          – DJohnM
          Dec 8 '18 at 22:17












          $begingroup$
          @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
          $endgroup$
          – gimusi
          Dec 8 '18 at 22:22




          $begingroup$
          @DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
          $endgroup$
          – gimusi
          Dec 8 '18 at 22:22












          $begingroup$
          Yes, I agree...
          $endgroup$
          – DJohnM
          Dec 8 '18 at 22:44




          $begingroup$
          Yes, I agree...
          $endgroup$
          – DJohnM
          Dec 8 '18 at 22:44


















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