anonymous
  • anonymous
calculus 1 or AP Calc - please guys help me Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^2, x = 4, and y = 1 about the y-axis.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
were you able to graph the equations and determine what the region looks like?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
|dw:1449536679259:dw|
anonymous
  • anonymous
this is how I solved it but I got it wrong
jim_thompson5910
  • jim_thompson5910
this is what the region is supposed to look like it's the region in blue see attached
1 Attachment
anonymous
  • anonymous
|dw:1449536783559:dw|
anonymous
  • anonymous
wait
anonymous
  • anonymous
how do you know the area that you must look for?
anonymous
  • anonymous
I've being doing the same mistake everytime
anonymous
  • anonymous
like you know the graphs make the region that I posted, how would I know which is the area that I should calculate the volume?
jim_thompson5910
  • jim_thompson5910
1 Attachment
zepdrix
  • zepdrix
Im pretty sure he had the correct region, yes Jim? They didn't list the x-axis as one of the boundaries.
jim_thompson5910
  • jim_thompson5910
oh that's true @zepdrix I assumed the x axis was thrown in there as well
anonymous
  • anonymous
okay, so for the graph y=4-x^2 and 2-x, the region that I need to find the volume is this one right? |dw:1449537161139:dw|
jim_thompson5910
  • jim_thompson5910
nvm my drawing then
anonymous
  • anonymous
okay, np, so is my integral correct?
jim_thompson5910
  • jim_thompson5910
you had the correct drawing the first time the correct shaded region is in green
1 Attachment
jim_thompson5910
  • jim_thompson5910
your integral is not correct
anonymous
  • anonymous
I got it from the formula dV= pi*(r2 - r1)^2 dy pi (4- x^(1/2) ) ^2 dy
anonymous
  • anonymous
can you tell me why
jim_thompson5910
  • jim_thompson5910
`pi (4- x^(1/2) ) ^2 dy` you have x in there, but the dy tells me that you're integrating with respect to y
jim_thompson5910
  • jim_thompson5910
you have the right idea, but it's slightly flawed
anonymous
  • anonymous
|dw:1449537600167:dw|
jim_thompson5910
  • jim_thompson5910
imagine you revolved the blue region around the y axis
1 Attachment
jim_thompson5910
  • jim_thompson5910
The volume of the blue solid (formed by the blue area revolving around the y axis) would be represented by this integral \[\Large \int_{1}^{16} \pi*(\sqrt{y})^2dy\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
so wee need to find the volume of the total green area right?
anonymous
  • anonymous
and then subtract the blue
jim_thompson5910
  • jim_thompson5910
the blue region+green region form to one big rectangle spin that around the y axis and you get a cylinder, which is represented by this integral \[\Large \int_{1}^{16}\pi*(4)^2dy\]
jim_thompson5910
  • jim_thompson5910
spinning around just the green region gives you a solid that has a (green solid volume) = (blue+green solid volume) - (blue solid volume) \[\large \text{green solid volume} = \left(\int_{1}^{16}\pi*(4)^2dy\right) - \left( \int_{1}^{16}\pi*(\sqrt{y})^2dy\right)\]
jim_thompson5910
  • jim_thompson5910
you can do a bit of simplification to get \[\Large \pi\int_{1}^{16}(16 - y)dy\]
anonymous
  • anonymous
112.5 pi
jim_thompson5910
  • jim_thompson5910
or \[\Large \frac{225\pi}{2}\]
anonymous
  • anonymous
okay, thank you. Could you help me with this last question, I know how to solve it but can you tell me what does it mean by perpendicular to the x-axis
anonymous
  • anonymous
The base of a solid in the region bounded by the parabola x2 + y = 4 and the line x + y = 2. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?
anonymous
  • anonymous
does it mean that it has to look like this
anonymous
  • anonymous
|dw:1449538253699:dw|
jim_thompson5910
  • jim_thompson5910
yep `perpendicular to the x-axis` means that each cross section is vertical like you drew just now
anonymous
  • anonymous
|dw:1449538398871:dw|
anonymous
  • anonymous
could I use that formula, oh i forgot my integral
anonymous
  • anonymous
|dw:1449538559611:dw|
jim_thompson5910
  • jim_thompson5910
one moment
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
ok I made a geogebra interactive applet http://ggbtu.be/miPl5o7r3 move point D and you'll see how the cross sections change notice how G is the midpoint of EF. You'll need to find the distance from G to F in terms of x (where x is restricted on the interval -1 <= x <= 2) to get the radius of a general semi-circle
anonymous
  • anonymous
okay I see
anonymous
  • anonymous
so that means that to find the diameter in the semicircle I would have to subtract the top function by the bottom?
jim_thompson5910
  • jim_thompson5910
anyways, after doing a bunch of work, you should find that the distance from F to G is represented by this expression \[\Large \frac{-x^2+x+2}{2}\] which is exactly equal to the radius of each semi-circle
anonymous
  • anonymous
|dw:1449540092155:dw|
jim_thompson5910
  • jim_thompson5910
yes, then divide by 2
jim_thompson5910
  • jim_thompson5910
to get the radius
anonymous
  • anonymous
okay. I understand now. Thank you so much for your time. I really hope you the best
jim_thompson5910
  • jim_thompson5910
you're welcome

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