# Thread: Revolving Area and Related Rates

1. ## Revolving Area and Rate of Change

I have three problems that I need help with,

1. $f(x)=-x^3+4x$
The area under the curve from $x=0$ to $x=2$ is revolved about the x-axis. Calculate the volume of the surface of revolution.
I used the disk method and found the volume to be $\frac{1024\pi}{105}$, can somone verify please?

2. $x^3y^4+xy^5=2$
Suppose an object is moving along with this curve. When the object arrived at (1,1) it was moving up with a speed of 4m/s. Calculate the speed and direction of the motion horizontally.
As well I have not had a problem like this before. I found the slope though to be $-4/9$ and the equation of the line to be $y=-4x/9+13/9$. I suppose i am to either derive the equation or integrate, but i don't have a clue.

3. A boy is 20m away from a rising balloon. The angle of elevation is increasing at a rate of 3 degrees/sec. At the instant the balloon is 8m above the ground find its speed of ascent.
d = 20m
$d\theta/dt$ = 3 deg/sec
$dh/dt=$? when h = 8m

I do not know what equation i would be using, i tried using $tan\theta=h/20$ and my answer was 69.6m/s but i think that is wrong.

Help appreciated very much.

2. I'll get you started with the first problem. The method is called Disk Method and we use it whenever the region that's being revolved is adjacent to the rotational axis.

The formula is INTEGRAL Pi (RADIUS^2) dx from x=a to x=b.

The RADIUS is the distance from the curve to the rotational axis, in your case it is simply the y-coordinate, so R(x) = -x^3 + 4x

Set up the integral from x=0 to x=2.

Good luck!

3. Originally Posted by apcalculus
I'll get you started with the first problem. The method is called Disk Method and we use it whenever the region that's being revolved is adjacent to the rotational axis.

The formula is INTEGRAL Pi (RADIUS^2) dx from x=a to x=b.

The RADIUS is the distance from the curve to the rotational axis, in your case it is simply the y-coordinate, so R(x) = -x^3 + 4x

Set up the integral from x=0 to x=2.

Good luck!
$\int\pi r^2dx$ from x=2 to x=0

Are you saying the radius = $-x^3 + 4x$?

4. Originally Posted by Raj
$\int\pi r^2dx$ from x=2 to x=0

Are you saying the radius = $-x^3 + 4x$?
Correct, so you're squaring the cubic function.

Some low-quality demos: DISK METHOD DEMO GALLERY

5. Hello, Raj!

Here's the second one . . .

2. . $x^3y^4 + xy^5 \:=\:2$
Suppose an object is moving along this curve.
When the object arrived at (1,1) it was moving up with a speed of 4 m/s.
Calculate the speed and direction of the motion horizontally.
Since this is a Related Rate problem,
. . you might expect to differentiate with respect to time.

When we do, we get: . $x^3\!\cdot\!4y^3\frac{dy}{dt} + 3x^2\!\cdot\!y^4\frac{dx}{dt} + x\!\cdot\!5y^4\frac{dy}{dt} + 1\!\cdot\!y^5\frac{dx}{dt} \:=\:0$

Re-arrange terms: . $3x^2y^4\frac{dx}{dt} + y^5\frac{dx}{dt} \;=\;-4x^3y^3\frac{dy}{dy} - 5xy^4\frac{dy}{dt}$

Factor: . $y^4(3x^2+y)\frac{dx}{dt} \;=\;-xy^3(4x^2 + 5y)\frac{dy}{dt}$

. . Hence: . $\frac{dx}{dt}\;=\;-\frac{x(4x^2+5y)}{y(3x^2+y)}\!\cdot\!\frac{dy}{dt}$ .[1]

We are given the point (1,1) and the object is moving up at 4 m/sec.
. . That is: . $x = 1,\;y = 1,\;\frac{dy}{dt} = 4$

Substitute into [1]: . $\frac{dx}{dt} \;=\;-\frac{1(4+5)}{1(3+1)}(4) \;=\;-9$

The object is moving at 9 m/sec to the left.

6. Hello again, Raj!

3. A boy is 20 m away from a rising balloon.
The angle of elevation is increasing at a rate of 3 degrees/sec.
At the instant the balloon is 8 m above the ground, find its speed of ascent.
Code:
                  *
* |
*   |
*     | h
*       |
* θ       |
* - - - - - *
20

We have: . $\tan\theta \:=\:\frac{h}{20} \quad\Rightarrow\quad h \:=\:20\tan\theta$

Differentiate with respect to time: . $\frac{dh}{dt} \:=\:20\!\cdot\!\sec^2\!\theta\!\cdot\!\frac{d\the ta}{dt}$ .[1]

We are given: . $\frac{d\theta}{dt} \:=\:3\text{ deg/sec} \:=\:\frac{\pi}{60}\text{ radians/sec}$

We are told that $h = 8$.
The right triangle looks like this:
Code:
                      *
*   |
*       | 8
* θ         |
* - - - - - - - *
20

We have: . $opp = 8,\;adj = 20$
Pythagorus says: . $hyp = \sqrt{464} = 4\sqrt{29}$
Hence: . $\sec\theta \:=\:\frac{hyp}{adj} \:=\:\frac{4\sqrt{29}}{20} \:=\:\frac{\sqrt{29}}{5}$

Substitute into [1]: . $\frac{dh}{dt}\;=\;20\left(\frac{\sqrt{29}}{5}\righ t)^2\!\left(\frac{\pi}{60}\right) \;=\;\frac{29\pi}{75}\text{ m/sec}$