• Mar 17th 2019, 11:40 AM
bigmansouf
Question: the area enclosed by $\displaystyle y=x^2-6x+18$ and y =10 is rotated about the x-axis; find the volume generated. take an element of area parallel to the x-axis of length$\displaystyle (x_{2}-x_{1})$ express the typical elements of volume in terms of y by using the fact that $\displaystyle x_{1}$ and $\displaystyle x_{2}$ are the roots of $\displaystyle x^2 -6x + (18-y)= 0$

my attempt:
the graph is Attachment 39299

$\displaystyle y=x^2-6x+18$
when x is the subject
$\displaystyle x = 3+(y-9)^{0.5}$

the interval is y =9 to y=10
the minimum point is (3,9)

$\displaystyle v= \int_{9}^{10} \pi (3+(y-9)^{0.5})^{2} \mathrm{d}y$
$\displaystyle v = 13.5000002 \pi$

please help me with this i am finding it difficult to understand how to tackle the book gives the answer as $\displaystyle 8\pi$
• Mar 17th 2019, 02:07 PM
Walagaster
Quote:

Question: the area enclosed by $\displaystyle y=x^2-6x+18$ and y =10 is rotated about the x-axis; find the volume generated. take an element of area parallel to the x-axis of length$\displaystyle (x_{2}-x_{1})$ express the typical elements of volume in terms of y by using the fact that $\displaystyle x_{1}$ and $\displaystyle x_{2}$ are the roots of $\displaystyle x^2 -6x + (18-y)= 0$

my attempt:
the graph is Attachment 39299

$\displaystyle y=x^2-6x+18$
when x is the subject
$\displaystyle x = 3+(y-9)^{0.5}$

There are two values of $x$ for each $y$
$$x=3 \pm \sqrt{y-9}$$

Quote:

the interval is y =9 to y=10
the minimum point is (3,9)

$\displaystyle v= \int_{9}^{10} \pi (3+(y-9)^{0.5})^{2} \mathrm{d}y$
$\displaystyle v = 13.5000002 \pi$

please help me with this i am finding it difficult to understand how to tackle the book gives the answer as $\displaystyle 8\pi$
So the length of your $dy$ element is $x_{right}-x_{left}$, using both $x$ values. Try that.
• Mar 17th 2019, 02:08 PM
MarkFL
First, we want to find the limits:

$\displaystyle x^2-6x+18=10$

$\displaystyle x^2-6x+8=0$

$\displaystyle (x-2)(x-4)=0$

And so, using the washer method:

$\displaystyle V=\pi\int_2^4 10^2-(x^2-6x+18)^2\,dx=\pi\int_2^4 -x^4+12x^3-72x^2+216x-224\,dx=\frac{128\pi}{5}$

Using the shell method, we find:

$\displaystyle V=4\pi\int_9^{10} y\sqrt{y-9}\,dy$

Let $\displaystyle u=y-9\implies du=dy$ and we have:

$\displaystyle V=4\pi\int_0^{1} (u+9)\sqrt{u}\,du=\frac{128\pi}{5}$
• Mar 17th 2019, 02:13 PM
Idea
we need

$\displaystyle x_2=3+\sqrt{-9+y}$

$\displaystyle x_1=3-\sqrt{-9+y}$

$\displaystyle x_2-x_1=2 \sqrt{-9+y}$

and so the element of volume is given by

$\displaystyle 2\pi y\left( 2 \sqrt{-9+y} \right)$
• Mar 17th 2019, 02:19 PM
Walagaster
Heh, guess I should have looked closer at what the OP did, apparently mixing up the two methods.
• Mar 24th 2019, 07:48 PM
bigmansouf
Quote:

First, we want to find the limits:

$\displaystyle x^2-6x+18=10$

$\displaystyle x^2-6x+8=0$

$\displaystyle (x-2)(x-4)=0$

And so, using the washer method:

$\displaystyle V=\pi\int_2^4 10^2-(x^2-6x+18)^2\,dx=\pi\int_2^4 -x^4+12x^3-72x^2+216x-224\,dx=\frac{128\pi}{5}$

Using the shell method, we find:

$\displaystyle V=4\pi\int_9^{10} y\sqrt{y-9}\,dy$

Let $\displaystyle u=y-9\implies du=dy$ and we have:

$\displaystyle V=4\pi\int_0^{1} (u+9)\sqrt{u}\,du=\frac{128\pi}{5}$

I have watched a video on youtube - https://www.youtube.com/watch?v=BDmlottZVd4 and it says that if it asks you about y-axis use $\displaystyle v = \int_{a}^{b}2 \pi xf(x)\mathrm{d} x$

thank you for the help
i went back and attempted it again

$\displaystyle y= x^2-6x+18$ and $\displaystyle y = 10$

$\displaystyle y= x^2-6x+18 = 10$
$\displaystyle y= x^2-6x+18 - 10 =0$
$\displaystyle y= x^2-6x+8 = 0$
thus x =4 and x =2
$\displaystyle v = \int_{2}^{4}2 \pi x(x^2-6x+8)\mathrm{d} x$
$\displaystyle v = \int_{2}^{4}2 \pi (x^3-6x^2+8x)\mathrm{d} x$
$\displaystyle v = 2 \pi \left [ \frac{x^{4}}{4}-2x^3+4x^2 \right ]_{2}^{4}$
$\displaystyle v= \left ( \left ( \frac{4^4}{4}-\left ( 2 \times 4^3 \right )+\left ( 4 \times4^2 \right ) \right )- \left ( \frac{2^4}{4} - \left ( 2 \times 2^3 \right )+ \left ( 4 \times 2^2 \right ) \right ) \right ) = - 8\pi$
but since volume is a positive quantity thus $\displaystyle v = 8 \pi$

please tell me is this a better approach and solution
• Mar 25th 2019, 06:36 AM
HallsofIvy
The fact that you got a negative number for volume should worry you! In the original problem you were told to "take an element of area parallel to the x-axis of length". But your "$\displaystyle x^2- 6x+ 18- 10$", the difference of two y values, is parallel to the y-axis, not the x-axis. In the u-tube video they used a formula for rotation around the y-axis, not the x- axis.
You have already been told that, using the "method of shells" you need the "shell" to be parallel to the axis of rotation. Here the axis of rotation is the x-axis, not the y-axis, so your "shell", parallel to the y-axis, has two x values at the end points. You have already been told that, since $\displaystyle y= x^2- 6x+ 18, x^2- 6x+ 18- y= 0$. $\displaystyle (x^2- 6x+ 9)+ 9- y= (x- 3)^2+ 9- y= 0$ so $\displaystyle (x- 3)^2= y- 9$, $\displaystyle x- 3= \pm \sqrt{y- 9}$ and $\displaystyle x= 3\pm\sqrt{y- 9}$. The length of a line parallel to the x-axis at height "y" across the parabola is $\displaystyle (3+ \sqrt{y- 9})- (3- \sqrt{y- 9})= 2\sqrt{y- 9}$. The axis of rotation, along the y- axis is y so rotates through a circle of circumference $\displaystyle 2\pi y$. the area of that "shell"" is $\displaystyle (2\pi y)(2\sqrt{y- 9})= 4\pi y \sqrt{y- 9}$. The "thickness" of each shell, again parallel to the y-axis, can be taken to be dy so the integration to find the volume is $\displaystyle 4\pi \int (y \sqrt{y- 9})dy$, integrating with respect to y, not x.