Volume of a solid

[JJ007]

Find the volume of a solid formed by revolving the region bounded by the graphs of $\displaystyle f(x)=2\sqrt{x} , x=4$, and the x-axis about the y-axis.

That's exactly how it's worded. Can someone point me in the right direction? How to start or which method?
Thanks.

 

[daon]

Washer method?

Bottom function: $\displaystyle f_B(x) = 0$
Top function: \(\displaystyle f_T(x) = 2\sqrt{x}}\)
Distance to axis: $\displaystyle r(x)=x$

Use the following formula:

$\displaystyle 2\pi \int_{x_a}^{x_b}r(x) \left [(f_T-f_B)\circ (x) \right ]dx$

In this case, we have

\(\displaystyle 2 \pi \int_{x_a = 0}^{x_b = 4} \left r(x)[(f_T-f_B)\circ (x) \right ]dx = 2 \pi \int_{0}^{4}x \left[ 2\sqrt{x}-0\right] dx\)

$\displaystyle = 4 \pi \int_{0}^{4}x^{3/2} dx$

 

[BigGlenntheHeavy]

$\displaystyle Shell:$

$\displaystyle 2\pi\int_{0}^{4} x(2\sqrt x)dx \ = \ \frac{256\pi}{5}$

$\displaystyle Disc:$

$\displaystyle \pi \int_{0}^{4}[4^{2}-(y^{2}/4)^{2}]dy \ = \ \frac{256\pi}{5}$

 

[daon]

Hah, I said washer, meant shell

 

[JJ007]

Ok, I understand the shell method. But I'm little off on the disk part. Here's a similar example I tried with disk:
revolving region bounded by graphs of $\displaystyle y=x^3, y=1,$ and $\displaystyle x=2$ about the y-axis.
$\displaystyle 1=x^3$
$\displaystyle x=1$
$\displaystyle v=\pi \int[R(x)]^2dx$
$\displaystyle v=\pi \int_1^2[x^3]^2dx = \frac{127}{7}\pi$

or
$\displaystyle v=\pi \int_1^2[x^3-1]^2dx$
$\displaystyle v=\pi \int_1^2[x^6-2x^3+1]dx=\frac{163}{14}\pi ?$

The real answer is supposed to be 120/7 pi

Thanks guys.

 

[daon]

Ok, I understand the shell method. But I'm little off on the disk part. Here's a similar example I tried with disk:
revolving region bounded by graphs of $\displaystyle y=x^3, y=1,$ and $\displaystyle x=2$ about the y-axis.
$\displaystyle 1=x^3$
$\displaystyle x=1$
$\displaystyle v=\pi \int[R(x)]^2dx$
$\displaystyle v=\pi \int_1^2[x^3]^2dx = \frac{127}{7}\pi$

or
$\displaystyle v=\pi \int_1^2[x^3-1]^2dx$
$\displaystyle v=\pi \int_1^2[x^6-2x^3+1]dx=\frac{163}{14}\pi ?$

The real answer is supposed to be 120/7 pi

Thanks guys.

Neither.

$\displaystyle R^2(x)-r^2(x)=(x^3)^2-1^2$.

$\displaystyle V=\pi \int_1^2 (x^6-1) dx$

 

[BigGlenntheHeavy]

$\displaystyle Note: \ This \ is \ revolving \ about \ the \ Y-AXIS.$

$\displaystyle Shell:$

$\displaystyle 2\pi\int_{1}^{2}(x^{3}-1)xdx \ = \ \frac{47\pi}{5}$

$\displaystyle Washer:$

$\displaystyle \pi\int_{1}^{8}(4-y^{2/3})dy \ = \ \frac{47\pi}{5}$

$\displaystyle See \ graph \ below:$

[attachment=0:jwq63sta]mno.jpg[/attachment:jwq63sta]

$\displaystyle This \ is \ revolving \ about \ the \ X-AXIS$

$\displaystyle Shell:$

$\displaystyle 2\pi\int_{1}^{8}y(2-y^{1/3})dy \ = \ \frac{120\pi}{7}$

$\displaystyle Washer:$

$\displaystyle \pi\int_{1}^{2}(x^{6}-1)dx \ = \ \frac{120\pi}{7}$