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anonymous
 4 years ago
Calculus: Rotating volume
anonymous
 4 years ago
Calculus: Rotating volume

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x = y^3, x = 4y y^3 = 4y y^3 4y = 0 y(y^2 4) = 0 y(y2)(y+2) = 0 they intersect at y= 0,2, 2 to graph: y = cubert(x), y = x/4 intersect at x = 0, 8, 8 dw:1328001094873:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm lol How wld u set up? Sorry I need giuded help :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0* set up the integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[V =\pi \int\limits_{8}^{8}R^{2} r^{2} dx\] so you are integrating over the circular areas of each crosssection R is outer radius from axis r is inner radius from axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh i see Thanks :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0welcome make sure you get R and r in terms of x, also this set up only works if you are revolving around a horizontal axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well it says the y axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just noticed something, because of the shape of the region, the outer and inner radius flip when x is negative. so just integrate from 0 to 8 but multiply by 2. due to symmetry the two regions will have identical volume
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