# Thread: Spinning around x axis

1. ## Spinning around x axis

Hi,

Im having trouble solving a really simple integration problem.

R is the finite region bounded by $x=y, x=4y-y^2$ are rotated around the x axis. Determine the volume of the solids.

Okay
The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

$A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2$

$=\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}$

But the answer is zero and it's wrong
What am i doing wrong?

2. Originally Posted by Jones
Hi,

Im having trouble solving a really simple integration problem.

R is the finite region bounded by $x=y, x=4y-y^2$ are rotated around the x axis. Determine the volume of the solids.

Okay
The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

$A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2$

$=\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}$

But the answer is zero and it's wrong
What am i doing wrong?
Not sure how you got an intersection at 4, nor the

$(4y-y^2-y)$ (Edit: no, this bit is fine, silly me!)

... maybe the two are connected? (Edit: no.)

Surely,

$x = y$ and $x = 4y - y^2$

give you $y = 0, 3$

? ...

3. I guess you are using shells to solve the porblem. I dont know how to use that method. However, tom@ballooncalculus is correct. The intersection points are:

(0,0) and (3,3)

because:

1. $x = y$
2. $x = 4y-y^2$
$y = 4y-y^2$
$y = y(4-y)$
$1 = 4-y$
$
y = 3
$

4. Originally Posted by Jones
Hi,

Im having trouble solving a really simple integration problem.

R is the finite region bounded by $x=y, x=4y-y^2$ are rotated around the x axis. Determine the volume of the solids.

Okay
The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

$A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2$

$=\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}$

But the answer is zero and it's wrong
What am i doing wrong?
You're using the disks method to evaluate this integral.

First you need to note that the graphs intersect at $y = 0$ and $y = 3$. So your LOWER limit is 0 and your UPPER limit is 3.

You also need to note that $4y - y^2 \geq y$ in the range $0 \leq y \leq 3$.

So the integral you need to set up is...

$\pi \int_0^3{(4y - y^2)^2 - y^2\,dy}$

$= \pi \int_0^3{16y^2 - 8y^3 + y^4 - y^2\,dy}$

$= \pi \int_0^3{y^4 - 8y^3 + 15y^2\,dy}$

$= \pi \left[\frac{1}{5}y^5 - 2y^4 + 5y^3\right]_0^3$

$= \pi \left[\frac{243}{5} - 162 + 135\right]$

$= \pi \left[\frac{108}{5}\right]$

$= \frac{108\pi}{5}\,\textrm{units}^3$

5. Originally Posted by Prove It
You're using the disks method to evaluate this integral.
I didn't think s/he was... Seems to me we're integrating perpendicular to the axis of revolution... Solid of revolution - Wikipedia, the free encyclopedia

____________________________________________

Don't integrate - balloontegrate!

Balloon Calculus Forum

6. Hello, Jones!

$R$ is the region bounded by $x\,=\,y,\;x\,=\,4y-y^2$ is rotated about the x axis.
Determine the volume of the solid.
Code:
          |
*               *
|       *     *
|           o(3,3)
|         *::::*
|       *::::::::o(4,2)
|     *::::::::*
|   *:::::::*
| *:::::*
- - o - - - - - - - - - - - - - -
|

Intersection: . $y \:=\:4y - y^2 \quad\Rightarrow\quad y^2-3y\:=\:0 \quad\Rightarrow\quad y(y-3)\:=\:0$

. . The graphs intersect at (0,0) and (3,3).

"Shells" Formula: . $V \;=\;2\pi\int^b_a y\left(x_{\text{right}} - x_{\text{left}}\right)\,dy$

We have: . $V \;=\;2\pi\int^3_0y\bigg[(4y-y^2) - y\bigg]\,dy \;=\;2\pi\int^3_0\bigg[3y^2 - y^3\bigg]\,dy$

Now finish up . . .

7. Originally Posted by Prove It
You're using the disks method to evaluate this integral.

First you need to note that the graphs intersect at $y = 0$ and $y = 3$. So your LOWER limit is 0 and your UPPER limit is 3.

You also need to note that $4y - y^2 \geq y$ in the range $0 \leq y \leq 3$.

So the integral you need to set up is...

$\pi \int_0^3{(4y - y^2)^2 - y^2\,dy}$

$= \pi \int_0^3{16y^2 - 8y^3 + y^4 - y^2\,dy}$

$= \pi \int_0^3{y^4 - 8y^3 + 15y^2\,dy}$

$= \pi \left[\frac{1}{5}y^5 - 2y^4 + 5y^3\right]_0^3$

$= \pi \left[\frac{243}{5} - 162 + 135\right]$

$= \pi \left[\frac{108}{5}\right]$

$= \frac{108\pi}{5}\,\textrm{units}^3$
But he is rotating about the x-axis, not the y-axis, so, how can you solve it with the disks method and still integrate along the y-axis?

Soroban's solution looks correct.