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anonymous

  • 5 years ago

Find the area of the solid obtained by rotating the region bounded by the given curves about the specified line. y=x^3, y=x, x>0

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  1. campbell_st
    • 5 years ago
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    find the point of intersections of the 2 curves....by equating the curves.... i.e \[x = x^3\] or \[x^3 - x =0 \] which gives \[x(x^2 -1)=0\] the points of intersection will be x = 0 and x = 1 (x = -1 not considered because of initial conditions. then the problem is \[V = \pi \int\limits_{0}^{1} (x^3 - x)^2 dx\]

  2. anonymous
    • 5 years ago
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    i thought you took the larger radius - the smaller radius. \[(x-x ^{3})\]?

  3. nikvist
    • 5 years ago
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    area, not volume !!!

  4. campbell_st
    • 5 years ago
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    The area between is rotated.... |dw:1327394267382:dw| the shape is like the horn of a trumpet if in doubt... include absolute value symbols around the integral...

  5. nikvist
    • 5 years ago
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    rotation about line y=x

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