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anonymous
 4 years ago
**Calc2 Help Needed**
Find the volume of the solid generated by revolving the described region about the given axis:
The region bounded above by the line y=6 , below by the curve y=sqrt(x), and to the left by the yaxis, rotated along the following lines:
x=36
anonymous
 4 years ago
**Calc2 Help Needed** Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=6 , below by the curve y=sqrt(x), and to the left by the yaxis, rotated along the following lines: x=36

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep.. I'm going to start tutoring next week. I am having the worst time with these..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try drawing out the region to get the bounds

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327644073706:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the bounds should be in y? so 0 to 6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was thinking of using the cylindrical shell method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02pi integral (radius)height

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0radius = y height = y36?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not as familiar with the shell method, mostly because i don't like using it anyway, wouldn't the radius be in terms of x, and the height be 6y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we would use the disc method?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so radius = y^2 and height is 6y?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you could use either i believe for this one. if you used the disc method: outer radius = 36 inner radius = 36y^2 ...yes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we have the integral of pi(3636+y^2)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0think area of circle: A = pi*r^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what should you integrate from, or what are the bounds

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the outer radius is 36 right, that is in xdirection now imagine the figure being sliced horizontally so we integrate over y what are the bounds of y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 0 to 6 in terms of y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so from 0 to 6 integral (36  36 + x^2)^2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not quite its R^2  r^2 Not (Rr)^2 > 36^2  (36y^2)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because you're not subtracting the radius, you are subtracting the areas

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh.. do I have two separate integrals?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no just keep them separate...you can simplify it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it right!! yay.. haha.. =3628.8pi. Thank you so much!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now it's time to study biology... fun :/
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