Math Help - Arc length and volume of the solid.

1. Arc length and volume of the solid.

Hello,
plz try to solve these questions.

2. Originally Posted by m777
Hello,
plz try to solve these questions.
Question #1:
"Find the volume of the solid that results when the region enclosed by
y = x^2 -1
x = 2
y = 0
is revolved about the y-axis."

Imagine, or draw the figure on paper.
The intersection of the vertical line x=2 and the parabola y = x^2 -1 is point (2,3).
The intersection of the horizontal line y=0 and the same parabola is point (1,0)

There are many ways to solve this.
One is by shell method, where the differential is dx.
One is by disc, washer, method, where the differential is dy
One is by parts. Volume of cylinder minus volume of paraboloid.

Let's do the shell method.
dV = (2pi*x)[(x^2 -1) -0](dx)
dV = 2pi(x^3 -x)dx --------------**

V = (2pi)INT.(1 to 2)[x^3 -x]dx
V = (2pi)[(x^4)/4 -(x^2)/2]|(1 to 2)
V = (2pi)[(16/4 -4/2) -(1/4 -1/2)]
V = (2pi)[2 +1/4]
V = 9pi/2 = 4.5pi cu.units ---------------answer.

---------------------------
The washer method.
dV = pi(2^2 -x^2)dy --------(i)

y = x^2 -1
So, x^2 = y+1
Substitute that in (i),
dV = pi(4 -(y+1))dy
dV = pi(3 -y)dy

Integrate both sides,
V = (pi)INT.(0 to 3)[3 -y]dy
V = (pi)[3y -(y^2)/2]|(0 to 3)
V = (pi)[(9 -9/2) -(0)]
V = 9pi/2 = 4.5pi cu.units ---------------same as above.

---------------------------------------------------------------
Question #2:
"Find the arc length of the curve
24xy = y^4 +48
from y=2 to y=4.

The usual dL = sqrt[1 +(dy/dx)^2]dx will not work easily because it is difficult to express the y in terms of x.
The x in terms of y is easy. So we use
dL = sqrt[1 +(dx/dy)^2]dy ----------------(1)

(1) is derived the same way the usual formula is done.
(dL)^2 = (dx)^2 +(dy)^2
Take the square roots of both sides,
dL = sqrt[(dx)^2 +(dy)^2]
Multiply the RHS by (dy/dy),
dL = sqrt[(dx/dy)^2 +(dy/dy)^2]dy
Hence,
dL = sqrt[1 +(dx/dy)^2]dy ----------**

24xy = y^4 +48
Differentiate both sides with respect to y,
24[x +y(dx/dy)] = 4y^3
24x +24y(dx/dy) = 4y^3
6x +6y(dy/dx) = y^3
(dx/dy) = (y^3 -6x)/(6y) ---------(i)

From (1),
x = (y^4 +48)/(24y)
Substitute that into (i),
dx/dy = (y^3 -6[(y^4 +48)/(24y)])/(6y)
dx/dy = (y^3 -(y^4 +48)/(4y)) /(6y)
dx/dy = (4y^4 -y^4 -48)/(24y^2)
dx/dy = (3y^4 -48)/(24y^2)
dx/dy = (y^4 -16)/(8y^2) ----------------(ii)

So, substitute that into (1),
dL = sqrt[1 +{(y^4 -16)/(8y^2)}^2]dy
dL = sqrt[{(64y^4) +(y^4 -16)^2} /(64y^4)]dy
dL = sqrt[{(64y^4) +(y^8 -32y^4 +256)} /(64y^4)]dy
dL = sqrt[(y^8 +32y^4 +256)/ (64y^4)]dy
dL = sqrt[(y^4 +16)^2 /(64y^4)]dy
dL = [(y^4 +16) /(8y^2)]dy
dL = [(y^2)/8 +16y^(-2)]dy ---------------(2)

Integrate both sides,
L = INT.(2 to 4)[(y^2)/8 +16y^(-2)]dy
L = [(y^3)/24 -16y^(-1)]|(2 to 4)
L = [(y^3)/24 -16/y]|(2 to 4)
L = [64/24 -16/4] -[8/24 -16/2]
L = [8/3 -4] -[1/3 -8]
L = 7/3 +4
L = 19/3 unit long -----------------answer.

3. #1:

${\pi}\int_{0}^{3}[2^{2}-(y+1)]dy$

= ${\pi}\int_{0}^{3}(3-y)dy$

4. Hello, m777!

1) Find the volume of the solid that results when the region enclosed by:
$y \:=\:x^2-1,\;x =2,\;y = 0$ is revolved about the y-axis.
A sketch is always welcome . . .
Code:
          |               *
|
|              *
|             *|
|           *::|
----+--------*-----+--
|    *   1     2
*
|
"Shells" formula: . $V \;=\;2\pi\int^b_axy\,dx$

We have: . $V \;=\;2\pi\int^2_1x(x^2 - 1)\,dx \;=\;2\pi\int^2_1\left(x^3 - x\right)\,dx \;= \;2\pi\left(\frac{x^4}{4} - \frac{x^2}{2}\right)\bigg]^2_1$

Evaluate: . $V \:=\:2\pi\left[\left(\frac{16}{4} - \frac{4}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right)\right] \:=\:2\pi\left(4 - 2 - \frac{1}{4} + \frac{1}{2}\right) \:=\:\boxed{\frac{9\pi}{2}}$

2) Find the arc length of the curve $24xy\:=\:y^4 + 48$ from $y = 2$ to $y = 4$
Since the problem is in terms of $y$, we'll use: . $L \:=\:\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy$

We have: . $x\:=\:\frac{y^4 + 48}{24y} \:=\:\frac{1}{24}y^3 + 2y^{-1}$

Then:. $\frac{dx}{dy}\:=\:\frac{1}{8}y^2 - 2y^{-2}$

. . $\left(\frac{dx}{dy}\right)^2\:=\:\left(\frac{1}{8} y^2 - 2y^{-2}\right)^2 \:=\:\frac{1}{64}y^4 - \frac{1}{2} + 4y^{-4}$

. . $1 + \left(\frac{dx}{dy}\right)^2\:=\:1 + \frac{1}{64}y^4 - \frac{1}{2} + 4y^{-4} \:=\:\frac{1}{64}y^4 + \frac{1}{2} + 4y^{-4}\;=\;\left(\frac{1}{8}y^2 + 2y^{-2}\right)^2$

. . $\sqrt{1 + \left(\frac{dx}{dy}\right)^2} \:=\:\frac{1}{8}y^2 + 2y^{-2}$

So we have: . $L \;=\;\int^4_2\left(\frac{1}{8}y^2 + 2y^{-2}\right)\,dy \;=\;\frac{1}{24}y^3 - 2y^{-1}\,\bigg]^4_2$

Evaluate: . $\left(\frac{1}{24}\cdot4^3 - \frac{2}{4}\right) - \left(\frac{1}{24}\cdot2^3 - \frac{2}{2}\right) \;=\;\left(\frac{64}{24} - \frac{1}{2}\right) - \left(\frac{8}{24} - 1\right)$

. . . . . . $= \;\frac{8}{3} - \frac{1}{2} - \frac{1}{3} + 1 \;=\;\boxed{\frac{5}{2}}$

But check my work . . . please!

5. Originally Posted by m777
Hello soroban,
i am very junior from you in age as well in knowlege so i cant check this
I think part of the reason he was asking you to check his work was to make sure you understood what he was doing. (The other part was likely to catch any stray errors he made.)

-Dan