Find the volume of the solid formed by rotating about the $x$-axis












4












$begingroup$


Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis



I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.



$$int_0^{0.7} pi (e^x+2)^2dx$$






calculus volume







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$endgroup$








  • 1




    $begingroup$
    The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
    $endgroup$
    – André Nicolas
    Feb 3 '14 at 19:37






  • 1




    $begingroup$
    Note you can pull the $pi$ out of the integral.
    $endgroup$
    – AlexR
    Feb 3 '14 at 19:37
















4












$begingroup$


Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis



I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.



$$int_0^{0.7} pi (e^x+2)^2dx$$






calculus volume







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
    $endgroup$
    – André Nicolas
    Feb 3 '14 at 19:37






  • 1




    $begingroup$
    Note you can pull the $pi$ out of the integral.
    $endgroup$
    – AlexR
    Feb 3 '14 at 19:37














4












4








4





$begingroup$


Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis



I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.



$$int_0^{0.7} pi (e^x+2)^2dx$$






calculus volume







share|cite|improve this question











$endgroup$




Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis



I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.



$$int_0^{0.7} pi (e^x+2)^2dx$$






calculus volume




calculus volume






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 '14 at 19:37









AlexR

22.8k12349




22.8k12349










asked Feb 3 '14 at 19:35









jannyjanny

951412




951412








  • 1




    $begingroup$
    The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
    $endgroup$
    – André Nicolas
    Feb 3 '14 at 19:37






  • 1




    $begingroup$
    Note you can pull the $pi$ out of the integral.
    $endgroup$
    – AlexR
    Feb 3 '14 at 19:37














  • 1




    $begingroup$
    The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
    $endgroup$
    – André Nicolas
    Feb 3 '14 at 19:37






  • 1




    $begingroup$
    Note you can pull the $pi$ out of the integral.
    $endgroup$
    – AlexR
    Feb 3 '14 at 19:37








1




1




$begingroup$
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
$endgroup$
– André Nicolas
Feb 3 '14 at 19:37




$begingroup$
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
$endgroup$
– André Nicolas
Feb 3 '14 at 19:37




1




1




$begingroup$
Note you can pull the $pi$ out of the integral.
$endgroup$
– AlexR
Feb 3 '14 at 19:37




$begingroup$
Note you can pull the $pi$ out of the integral.
$endgroup$
– AlexR
Feb 3 '14 at 19:37










3 Answers
3






active

oldest

votes


















0












$begingroup$

Your equation is correct.
$$begin{align*}
A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^{1.4} e^x dx + int_0^{0.7} e^x dx + 2.8right)\
& = pi left( frac12 e^{1.4} + e^{0.7} + 2.8 right) \
& approx 21.4927ldots
end{align*}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not sure your answer is correct. Note $e^0ne0$.
    $endgroup$
    – John Molokach
    Jan 9 '16 at 14:59












  • $begingroup$
    @JohnMolokach $e^0 = 1$ by definition...
    $endgroup$
    – AlexR
    Jan 9 '16 at 15:00










  • $begingroup$
    Sorry, edited...
    $endgroup$
    – John Molokach
    Jan 9 '16 at 15:01










  • $begingroup$
    It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
    $endgroup$
    – user334732
    May 17 '18 at 10:46












  • $begingroup$
    @RobertFrost That‘s a substitution, yielding the factor of $1/2$
    $endgroup$
    – AlexR
    May 17 '18 at 11:18



















0












$begingroup$

Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.



$$begin{align*}
A/pi & = int_0^{a} (e^x + c)^2 dx = int_0^{a} e^{2x} dx + 2c int_0^{a} e^x dx + c^2 a\
& = frac12 ( e^{2a}-1) + 2c(e^{a}-1) dx + ac^2.\
end{align*}$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer



    begin{align*}
    A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4int_0^{0.7}dxright)\
    & = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
    & = pi left( frac12 int_0^{1.4} e^x dx + 4int_0^{0.7} e^x dx + 2.8right)qquad qquad(u=2x,du=2dx)\
    & = pi left( frac12 left(e^{1.4}-e^0right) + 4left(e^{0.7}-e^0right) + 2.8 right) \
    & approx 26.3347ldots
    end{align*}






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Your equation is correct.
      $$begin{align*}
      A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
      & = pi left( frac12 int_0^{1.4} e^x dx + int_0^{0.7} e^x dx + 2.8right)\
      & = pi left( frac12 e^{1.4} + e^{0.7} + 2.8 right) \
      & approx 21.4927ldots
      end{align*}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Not sure your answer is correct. Note $e^0ne0$.
        $endgroup$
        – John Molokach
        Jan 9 '16 at 14:59












      • $begingroup$
        @JohnMolokach $e^0 = 1$ by definition...
        $endgroup$
        – AlexR
        Jan 9 '16 at 15:00










      • $begingroup$
        Sorry, edited...
        $endgroup$
        – John Molokach
        Jan 9 '16 at 15:01










      • $begingroup$
        It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
        $endgroup$
        – user334732
        May 17 '18 at 10:46












      • $begingroup$
        @RobertFrost That‘s a substitution, yielding the factor of $1/2$
        $endgroup$
        – AlexR
        May 17 '18 at 11:18
















      0












      $begingroup$

      Your equation is correct.
      $$begin{align*}
      A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
      & = pi left( frac12 int_0^{1.4} e^x dx + int_0^{0.7} e^x dx + 2.8right)\
      & = pi left( frac12 e^{1.4} + e^{0.7} + 2.8 right) \
      & approx 21.4927ldots
      end{align*}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Not sure your answer is correct. Note $e^0ne0$.
        $endgroup$
        – John Molokach
        Jan 9 '16 at 14:59












      • $begingroup$
        @JohnMolokach $e^0 = 1$ by definition...
        $endgroup$
        – AlexR
        Jan 9 '16 at 15:00










      • $begingroup$
        Sorry, edited...
        $endgroup$
        – John Molokach
        Jan 9 '16 at 15:01










      • $begingroup$
        It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
        $endgroup$
        – user334732
        May 17 '18 at 10:46












      • $begingroup$
        @RobertFrost That‘s a substitution, yielding the factor of $1/2$
        $endgroup$
        – AlexR
        May 17 '18 at 11:18














      0












      0








      0





      $begingroup$

      Your equation is correct.
      $$begin{align*}
      A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
      & = pi left( frac12 int_0^{1.4} e^x dx + int_0^{0.7} e^x dx + 2.8right)\
      & = pi left( frac12 e^{1.4} + e^{0.7} + 2.8 right) \
      & approx 21.4927ldots
      end{align*}$$






      share|cite|improve this answer









      $endgroup$



      Your equation is correct.
      $$begin{align*}
      A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
      & = pi left( frac12 int_0^{1.4} e^x dx + int_0^{0.7} e^x dx + 2.8right)\
      & = pi left( frac12 e^{1.4} + e^{0.7} + 2.8 right) \
      & approx 21.4927ldots
      end{align*}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Feb 3 '14 at 19:41









      AlexRAlexR

      22.8k12349




      22.8k12349












      • $begingroup$
        Not sure your answer is correct. Note $e^0ne0$.
        $endgroup$
        – John Molokach
        Jan 9 '16 at 14:59












      • $begingroup$
        @JohnMolokach $e^0 = 1$ by definition...
        $endgroup$
        – AlexR
        Jan 9 '16 at 15:00










      • $begingroup$
        Sorry, edited...
        $endgroup$
        – John Molokach
        Jan 9 '16 at 15:01










      • $begingroup$
        It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
        $endgroup$
        – user334732
        May 17 '18 at 10:46












      • $begingroup$
        @RobertFrost That‘s a substitution, yielding the factor of $1/2$
        $endgroup$
        – AlexR
        May 17 '18 at 11:18


















      • $begingroup$
        Not sure your answer is correct. Note $e^0ne0$.
        $endgroup$
        – John Molokach
        Jan 9 '16 at 14:59












      • $begingroup$
        @JohnMolokach $e^0 = 1$ by definition...
        $endgroup$
        – AlexR
        Jan 9 '16 at 15:00










      • $begingroup$
        Sorry, edited...
        $endgroup$
        – John Molokach
        Jan 9 '16 at 15:01










      • $begingroup$
        It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
        $endgroup$
        – user334732
        May 17 '18 at 10:46












      • $begingroup$
        @RobertFrost That‘s a substitution, yielding the factor of $1/2$
        $endgroup$
        – AlexR
        May 17 '18 at 11:18
















      $begingroup$
      Not sure your answer is correct. Note $e^0ne0$.
      $endgroup$
      – John Molokach
      Jan 9 '16 at 14:59






      $begingroup$
      Not sure your answer is correct. Note $e^0ne0$.
      $endgroup$
      – John Molokach
      Jan 9 '16 at 14:59














      $begingroup$
      @JohnMolokach $e^0 = 1$ by definition...
      $endgroup$
      – AlexR
      Jan 9 '16 at 15:00




      $begingroup$
      @JohnMolokach $e^0 = 1$ by definition...
      $endgroup$
      – AlexR
      Jan 9 '16 at 15:00












      $begingroup$
      Sorry, edited...
      $endgroup$
      – John Molokach
      Jan 9 '16 at 15:01




      $begingroup$
      Sorry, edited...
      $endgroup$
      – John Molokach
      Jan 9 '16 at 15:01












      $begingroup$
      It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
      $endgroup$
      – user334732
      May 17 '18 at 10:46






      $begingroup$
      It might be me, but are you sure you can make the simplification of taking $e^{2x}$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
      $endgroup$
      – user334732
      May 17 '18 at 10:46














      $begingroup$
      @RobertFrost That‘s a substitution, yielding the factor of $1/2$
      $endgroup$
      – AlexR
      May 17 '18 at 11:18




      $begingroup$
      @RobertFrost That‘s a substitution, yielding the factor of $1/2$
      $endgroup$
      – AlexR
      May 17 '18 at 11:18











      0












      $begingroup$

      Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.



      $$begin{align*}
      A/pi & = int_0^{a} (e^x + c)^2 dx = int_0^{a} e^{2x} dx + 2c int_0^{a} e^x dx + c^2 a\
      & = frac12 ( e^{2a}-1) + 2c(e^{a}-1) dx + ac^2.\
      end{align*}$$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.



        $$begin{align*}
        A/pi & = int_0^{a} (e^x + c)^2 dx = int_0^{a} e^{2x} dx + 2c int_0^{a} e^x dx + c^2 a\
        & = frac12 ( e^{2a}-1) + 2c(e^{a}-1) dx + ac^2.\
        end{align*}$$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.



          $$begin{align*}
          A/pi & = int_0^{a} (e^x + c)^2 dx = int_0^{a} e^{2x} dx + 2c int_0^{a} e^x dx + c^2 a\
          & = frac12 ( e^{2a}-1) + 2c(e^{a}-1) dx + ac^2.\
          end{align*}$$






          share|cite|improve this answer











          $endgroup$



          Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.



          $$begin{align*}
          A/pi & = int_0^{a} (e^x + c)^2 dx = int_0^{a} e^{2x} dx + 2c int_0^{a} e^x dx + c^2 a\
          & = frac12 ( e^{2a}-1) + 2c(e^{a}-1) dx + ac^2.\
          end{align*}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 25 '17 at 11:23

























          answered Apr 25 '17 at 9:39









          NarasimhamNarasimham

          21.2k62258




          21.2k62258























              0












              $begingroup$

              As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer



              begin{align*}
              A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4int_0^{0.7}dxright)\
              & = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
              & = pi left( frac12 int_0^{1.4} e^x dx + 4int_0^{0.7} e^x dx + 2.8right)qquad qquad(u=2x,du=2dx)\
              & = pi left( frac12 left(e^{1.4}-e^0right) + 4left(e^{0.7}-e^0right) + 2.8 right) \
              & approx 26.3347ldots
              end{align*}






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer



                begin{align*}
                A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4int_0^{0.7}dxright)\
                & = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
                & = pi left( frac12 int_0^{1.4} e^x dx + 4int_0^{0.7} e^x dx + 2.8right)qquad qquad(u=2x,du=2dx)\
                & = pi left( frac12 left(e^{1.4}-e^0right) + 4left(e^{0.7}-e^0right) + 2.8 right) \
                & approx 26.3347ldots
                end{align*}






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer



                  begin{align*}
                  A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4int_0^{0.7}dxright)\
                  & = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
                  & = pi left( frac12 int_0^{1.4} e^x dx + 4int_0^{0.7} e^x dx + 2.8right)qquad qquad(u=2x,du=2dx)\
                  & = pi left( frac12 left(e^{1.4}-e^0right) + 4left(e^{0.7}-e^0right) + 2.8 right) \
                  & approx 26.3347ldots
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$



                  As this just got bumped from the homepage and @AlexR' answer is not correct, here the correct answer



                  begin{align*}
                  A & = pi int_0^{0.7} (e^x + 2)^2 dx = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4int_0^{0.7}dxright)\
                  & = pi left( int_0^{0.7} e^{2x} dx + 4 int_0^{0.7} e^x dx + 4cdot0.7 right)\
                  & = pi left( frac12 int_0^{1.4} e^x dx + 4int_0^{0.7} e^x dx + 2.8right)qquad qquad(u=2x,du=2dx)\
                  & = pi left( frac12 left(e^{1.4}-e^0right) + 4left(e^{0.7}-e^0right) + 2.8 right) \
                  & approx 26.3347ldots
                  end{align*}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 10:53









                  Bo5manBo5man

                  518




                  518






























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