anonymous
  • anonymous
*CALC2 HELP PLEASE* The region enclosed by the curve x=sqrt(y), and by the lines x= -y and y=5 rotated around the x-axis.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Thanks for the medal, but do you have any tips?
anonymous
  • anonymous
well, what you've got yourself there is an integral with some pretty odd bounds, give me a second so I can draw a picture to see how to approach it.
anonymous
  • anonymous
Thanks man.. The shape is pretty funky..

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anonymous
  • anonymous
is it 3d?
anonymous
  • anonymous
|dw:1327642062679:dw|
anonymous
  • anonymous
yeah it's rotating
anonymous
  • anonymous
another, question, which area contained by those lines?
anonymous
  • anonymous
It doesn't specify really, so I'm confused there as well..
anonymous
  • anonymous
The whole area I shaded in
anonymous
  • anonymous
Because graph is this
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anonymous
  • anonymous
hmm.. that is really weird..
anonymous
  • anonymous
aww, I thought dumbcow had it figured out.. haha
anonymous
  • anonymous
so depending on what the area is its just a matter of doing the volume of revolution, but without knowing which specific area is... I want to assume that the area is the bit off to the left from y=-x up to y=5 and from y=-x over to y=x^2
anonymous
  • anonymous
did they specify x=+sqrt(y)?
dumbcow
  • dumbcow
ok i think you can approach this using the washer method where you have an outer area - inner area each cross section is a circle and you integrate wrt x from -5 to sqrt5 the outer radius would be 5 since that is the max y-value. the inner radius would be the y-value of the appropriate function f(x) |dw:1327642598764:dw| \[\large V = \pi \int\limits_{-5}^{0}(25-x^{2}) dx + \pi \int\limits_{0}^{\sqrt{5}}(25-x^{4}) dx\]
anonymous
  • anonymous
dumbcow: after integration I got a negative volume?
dumbcow
  • dumbcow
hmm no it should be positive make sure when you evaluate you subtract upper limit - lower limit
anonymous
  • anonymous
ok let me try again
anonymous
  • anonymous
the integration would be.. pi(25x-x^3/3) and pi(25x-x^5/5), right?
dumbcow
  • dumbcow
yes
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%5Bpi+integral+%5B%5B25-x%5E2%5D%2C+%7Bx%2C-5%2C+0%7D%5D%5D+%2B+%5Bpi+integral+%5B%5B25-x%5E4%5D%2C+%7Bx%2C0%2C+sqrt%285%29%7D%5D%5D
anonymous
  • anonymous
I got 402.296 on wolfram, and it says incorrect on my homework.. :/
anonymous
  • anonymous
Nahhh I'm just kidding.. Worked perfect. Thanks guys!!
dumbcow
  • dumbcow
isn't wolfram great. ok well do they want in terms of pi? i'll look at it again i may have made a mistake somewhere
dumbcow
  • dumbcow
never mind :)

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