anonymous
  • anonymous
What integral represents the volume of the solid formed by revolving the region bounded by the graphs of y = x^3 y = 1 and x = 2 about the line x = 2
Mathematics
schrodinger
  • schrodinger
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mLe
  • mLe
Hello @Yaros and Welcome to OpenStudy! Thanks for asking a Qualified Help Question first try graphing out what you have been given so that you have a visual representation of what you are looking for :)
Michele_Laino
  • Michele_Laino
your problem can be represented by this drawing: |dw:1435594893861:dw|
Michele_Laino
  • Michele_Laino
now I make this traslation: \[\Large \left\{ \begin{gathered} x = X + 2 \hfill \\ y = Y \hfill \\ \end{gathered} \right.\] where X, Y are the new coordinates, furthermore, we can see that the origin of the plane X-Y is lòocated at point x=2,y=0

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Michele_Laino
  • Michele_Laino
|dw:1435595055446:dw|
Michele_Laino
  • Michele_Laino
Now I refer the equation of our cubic parabola to the new reference system, so I get: \[\Large y = {x^3} \Rightarrow Y = {\left( {X - 2} \right)^3}\]
Michele_Laino
  • Michele_Laino
so the requested volume is given by the subsequent integral: \[\Large V = \pi \int_0^1 {dY\left\{ {{{\left( {\sqrt[3]{Y} - 2} \right)}^2} - {{\left( { - 1} \right)}^2}} \right\}} \]
Michele_Laino
  • Michele_Laino
oops.. the right equation of our paraboa using the coordinates X,Y is: \[\Large y = {x^3} \Rightarrow Y = {\left( {X + 2} \right)^3}\]
Michele_Laino
  • Michele_Laino
|dw:1435595559191:dw|
Michele_Laino
  • Michele_Laino
please keep in mind that we are considering the rotation, around the x=2 axis
Michele_Laino
  • Michele_Laino
please keep in mind that we are considering the rotation, around the x=2 axis
Michele_Laino
  • Michele_Laino
after a simple computation, we get: \[\Large V = \pi \left. {\left( {\frac{3}{5}{Y^{5/3}} + 3Y - 3{Y^{4/3}}} \right)} \right|_0^1 = \frac{{3\pi }}{5}\]
Michele_Laino
  • Michele_Laino
|dw:1435596051708:dw|
Michele_Laino
  • Michele_Laino
|dw:1435596146656:dw|
Michele_Laino
  • Michele_Laino
that integral is the volume generated by the rotation of the shaded region above
Michele_Laino
  • Michele_Laino
now we have to add the volume of the cylinder whose radius is r=1 and height =1
Michele_Laino
  • Michele_Laino
so we get: \[\Large {V_{TOTAL}} = V + \pi = \frac{{3\pi }}{5} + \pi = \frac{{8\pi }}{5}\]
Michele_Laino
  • Michele_Laino
Another way to compute the requested volume, is to apply the subsequent formula: \[\Large {V_{TOTAL}} = \pi {\int_0^1 {dY\left( {\sqrt[3]{Y} - 2} \right)} ^2}\]

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