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

$\displaystyle y=x^2-6x+18$
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 $\displaystyle 8\pi$

Originally Posted by 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

$\displaystyle y=x^2-6x+18$
when x is the subject
$\displaystyle x = 3+(y-9)^{0.5}$
There are two values of $x$ for each $y$
$$x=3 \pm \sqrt{y-9}$$

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.

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:

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

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

$\displaystyle 2\pi y\left( 2 \sqrt{-9+y} \right)$

Heh, guess I should have looked closer at what the OP did, apparently mixing up the two methods.

Originally Posted by 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:

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

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

This is extremely troubling! You got several responses here, pointing out what mistakes you had made and what you should do, but apparently did not understand them! You then watched a u-tube video and did not understand that!

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.