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anonymous
 4 years ago
*CALC2 HELP PLEASE* The region enclosed by the curve x=sqrt(y), and by the lines x= y and y=5 rotated around the xaxis.
anonymous
 4 years ago
*CALC2 HELP PLEASE* The region enclosed by the curve x=sqrt(y), and by the lines x= y and y=5 rotated around the xaxis.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for the medal, but do you have any tips?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks man.. The shape is pretty funky..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327642062679:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0another, question, which area contained by those lines?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It doesn't specify really, so I'm confused there as well..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The whole area I shaded in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because graph is this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm.. that is really weird..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aww, I thought dumbcow had it figured out.. haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0did they specify x=+sqrt(y)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 yvalue. the inner radius would be the yvalue of the appropriate function f(x) dw:1327642598764:dw \[\large V = \pi \int\limits_{5}^{0}(25x^{2}) dx + \pi \int\limits_{0}^{\sqrt{5}}(25x^{4}) dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dumbcow: after integration I got a negative volume?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm no it should be positive make sure when you evaluate you subtract upper limit  lower limit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the integration would be.. pi(25xx^3/3) and pi(25xx^5/5), right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got 402.296 on wolfram, and it says incorrect on my homework.. :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nahhh I'm just kidding.. Worked perfect. Thanks guys!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isn'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
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