Hey there, I've been having some trouble solving these. Any help would be appreciated.
Find volume of shape:
1. y=sqrt(x), with bounds y=0, x=4, rotated around x axis
2. y=sqrt(16-x^2), with bounds x=0, y=0, rotated around y axis
3. y=x, with bounds x=0, x=6, y=4, rotated around x=6
RealPro
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This is basic. You have read the examples yes? There will be no trouble if you draw the diagrams.
Consider the solid of revolution of the curve f(x) about x axis, from x=a to x=b.
If you have a plane perpendicular to the x axis, then you can define V(x) as the volume of the solid that lies to the "left" of the plane, i.e. at those x which are less than the x of the plane.
Then if you move the plane a very small distance dx along the x axis, what will be the added volume? We must know our geometry formulas
dV = pi * (f(x))^2 dx
Clearly the total volume of the solid is V(b) - V(a).
The Newton-Leibniz formula (fundamental theorem of calculus) says that you can get this by integrating the derivative dV/dx from a to b.
V_tot = int[a to b] (dV/dx) dx
V_tot = int[a to b] (pi * (f(x))^2) dx
Intuitively we think of this as adding up the volumes of all those infinitesimally thin disks of which the solid consists.
1.
Since this solid's leftmost point is at x=0 and rightmost at x=4, then
V = pi * int[0 to 4] (sqrtx)^2 dx = 8pi
2.
Here we integrate along y. The solid goes from y=0 to y=4.
x = f(y) = sqrt(16 - y^2)
pi * int[0 to 4] (f(y))^2 dy
3.
Ambiguous bounds. Ignoring the bound x=6,
V = pi * int[0 to 4] ( 6^2 - (6-y)^2 ) dy
Using the method of shells you get
1.
A horizontal slice through the region at y, when rotated, will produce a cylindrical shell of radius y and height (length) 4-y^2
2pi * int[0 to 2] y(4-y^2) dy
2.
A vertical slice at x produces a shell of radius x and height sqrt(16-x^2)
2pi * int[0 to 4] x sqrt(16-x^2) dx
3.
A vertical slice at x will produce a shell of radius 6-x and height 4-x.
2pi * int[0 to 4] (6-x)(4-x)dx
2 and 3 don't really require calculus if you can just use the formulas for a hemisphere and truncated cone though.
stanschim
1. π∫y^2 dx, where x runs from 0 to 4.
Transform this into: π∫x dx = π(x^2)/2 from x = 0 to 4. Answer: 8π cubic units.
2. π∫x^2 dy, where y runs from 0 to 4.
Transform this into: π∫ (16-y^2) dy, where y runs from 0 to 4. Answer: 64π cubic units.
3. π∫(6-x)^2 dy , where y runs from 0 to 4.
Transform this into π∫(6-y)^2 dy, where y runs from 0 to 4. Answer: 80π/3 cubic units.