Washers

Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
The given is :
lines: x = 4, y = 8
The curve is: y^2 = x^3
Question:
Find the volume of The region below the line y = 8 (meaning inside the parabola) revolving about the line x = 4


I cant hit the answer key and i always end up with 1024/35 pi cubic units ...

the answer is suupposed to be 3456/35 pi cubic units from the answer key

help! and thanks alot!

Hello, ^_^Engineer_Adam^_^!

Your sketch may be incorrect . . .

Quote:

---

Find the volume of the solid of revolution generated
when the given region is revolved about the indicated line.

Region bounded by: .y² = x³, .x = 0, .y = 8
. . revolved about x = 4.

Answer: .3456π/35 units³

---

Code:

---

        |
      8+ - - - - -*
        |::::::::::|
        |:::::::::*|
        |::::::::* |
        |::::::*  |
        |:::*      |
      --*----------+--
        |          4

---

I prefer to use Cylindrical Shells.

. . V .= .2π ∫ (radius)(height) dx

The radius is: .4 - x
The height is: .8 - x^{3/2}


We have: . V . = . 2π ∫ (4 - x)·(8 - x^{3/2}) dx

. . . . . . . . . . .= . 2π ∫ (32 - 8x - 4x^{3/2} + x^{5/2}) dx

Is there a way to use
the volumes by slicing and rotating method?

Find the volume of The region below the line y = 8 (meaning inside the parabola) revolving about the line x = 4

It is the thing inside the parabola.

Then the radius about the vertical line, the axis of rotation, x=4 is not a single radius. There are two radii involved---the inner radius and the outer radius. That is why you mentioned "washers".

inner radius = (4 -x)
outer radius = 4 -0 = 4

So your infinitesimal volume, or dV, is a washer whore radii are those, and whose thickness is dy.
Hence, dV = {pi[(outer r)^2] -pi[(inner r)^2]}*dy
dV = pi[(outer r)^2 -(inner r)^2]dy
dV = pi[4^2 -(4-x)^2]dy
dV = pi[16 -16 +8x -x^2]dy
dV = pi[-x^2 +8x]dy

But dy has nothing to do with x's in integration, so convert the x's into their equivalent y's.

y^2 = x^3
Raise both sides to their 1/3 powers,
y^(2/3) = x
And x^2 = y^(4/3)

So,
dV = pi[-y^(4/3) +8y^(2/3)]dy ----(i)

Boundaries for dy are y=0 and y=8

So, intregrating both sides of (i),
V = (pi)INT.(0-->8)[-y^(4/3) +8y^(2/3)]dy
V = (pi)[-(3/7)y^(7/3) +(24/5)y^(5/3]|(0-->8)
V = pi{[-(3/7)(8^(7/3)) +(24/5)(8^(5/3))] -[0]}
V = pi{-(3/7)(2^7) +(24/5)(2^5)}
V = pi{-384/7 +768/5}
V = pi(3456/35)
V = (3456/35)pi cu.units --------------answer.