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use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the...Asked by Neesh
Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.
y = 8x − x^2, y = x
y = 8x − x^2, y = x
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Answered by
oobleck
The curves intersect at (0,0) and (7,7)
So, using shells of thickness dx,
v = ∫[0,7] 2πrh dx
where r=x and h=(8x-x^2)-x = 7x-x^2
v = ∫[0,7] 2πx(7x-x^2) dx = 2401π/6
To use discs of thickness dy, you need to change the boundary at (7,7) so it gets a bit more complicated.
v1 = ∫[0,7] π(R^2-r^2) dy
where R = y and r = 4-√(16-y)
v1 = ∫[0,7] π(y^2-(4-√(16-y))^2) dy = 673π/6
v2 = ∫[7,16] π(R^2-r^2) dy
where R = 4+√(16-y) and r = 4-√(16-y)
v2 = ∫7,16] π*((4+√(16-y))^2-(4-√(16-y))^2) dy = 288π
so v = v1+v2 = 673π/6 + 1728π/6 = 2401π/6
So, using shells of thickness dx,
v = ∫[0,7] 2πrh dx
where r=x and h=(8x-x^2)-x = 7x-x^2
v = ∫[0,7] 2πx(7x-x^2) dx = 2401π/6
To use discs of thickness dy, you need to change the boundary at (7,7) so it gets a bit more complicated.
v1 = ∫[0,7] π(R^2-r^2) dy
where R = y and r = 4-√(16-y)
v1 = ∫[0,7] π(y^2-(4-√(16-y))^2) dy = 673π/6
v2 = ∫[7,16] π(R^2-r^2) dy
where R = 4+√(16-y) and r = 4-√(16-y)
v2 = ∫7,16] π*((4+√(16-y))^2-(4-√(16-y))^2) dy = 288π
so v = v1+v2 = 673π/6 + 1728π/6 = 2401π/6
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