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volume of solid of revolution - please help

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 $

Re: volume of solid of revolution - please help

Quote:

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

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.

Re: volume of solid of revolution - please help

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:

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

Re: volume of solid of revolution - please help

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

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

Re: volume of solid of revolution - please help

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

Re: volume of solid of revolution - please help

Quote:

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

Re: volume of solid of revolution - please help

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.