[Volumes] Question (1 Viewer)

[Pyrobooby]

The rectangle OABC, where O(0,0), A(a,0), C(0,c) is cut through by a continuous, smooth, differentiable curve y=f(x). The region below y=f(x) is rotated about he side OA while the region above y=f(x) is rotated about the side BC. If the volumes of the two solids are equal, show that the two areas are equal.

I'm not sure how to go about this question. I can see that the coordinate B will be B(a,c), and the two solids generated are like a bowl and a cone.

On a side note: is the integral of [f(x)]^2 equal to 1/3 [f(x)]^3 ?

ALSO!
Is there another way to derive the integration formula in Cambridge Ex. 6.1 Example 3? Because that is some whack ass shit right there. The question is: the region bounded by the curve y=x(4-x) and the x-axis is rotated about the y-axis. Find the volume of the solid of revolution by taking slices perpendicular to the y-axis. The author uses the sum and product of roots to derive the formula.

 

[Drongoski]

On a side note: is the integral of [f(x)]^2 equal to 1/3 [f(x)]^3 ?

Only if integrated with respect to f(x) rather than x.

 

[jyu]

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[Pyrobooby]

Thanks jyu, ingenious (Y)

 

[Trebla]

Is there another way to derive the integration formula in Cambridge Ex. 6.1 Example 3? Because that is some whack ass shit right there. The question is: the region bounded by the curve y=x(4-x) and the x-axis is rotated about the y-axis. Find the volume of the solid of revolution by taking slices perpendicular to the y-axis. The author uses the sum and product of roots to derive the formula.

You can use a symmetry argument here. Let x be the x₁ shown on that diagram. Draw a vertical line x = 4 and it should be clear by symmetry that
x₂ = 4 - x
Now from the original equation
x² - 4x = - y
=> (x - 2)² = 4 - y
=> x = 2 ± √(4 - y)
But since we defined x in the domain 0 < x < 2 (i.e. the left half of the curve) then we take the negative root, hence
x = 2 - √(4 - y)
We now obtain
dV = π((4 - x)² - x²)dy
= 8π(2 - x)dy
= 8π√(4 - y)dy

 

[Pyrobooby]

Thank you trebla!

One last question:
The lows of a point P on the rim of a circular wheel of unit radius rolling on a straight line is defined by:
x=(theta) - sin(theta)
y=1-cos(theta).

Find the volume of the solid formed by rotating the arch under the curve, for 0<=(theta)<=2pi, about the x-axis.

And another:
A torus is formed by rotating the circle x^2+y^2=a^2 about the line x=2a. Find the volume so generated. <== I arrived at the integral 16pi a (from y=0 to y=a) ((a^2-y^2)^(1/2)) dy. I used wolfram to integrate and it gave me the correct answer, but I can't seem to get there myself!