Volume Question - shell method (1 Viewer)

asb_92 · Jun 9, 2009

oki.. so i hav a test on volumes in 2 days.. and i dnt get how teh heckk i gta do this stufff >< can soemone plz help me out with this qu?

The area enclosed by y=x^2 and x=y^2 is rotated about y-axis. Find the volume.

thanks veryy muchh!
and also if u guys have any tips on how to approach volumes questions and stuff?

thanks again

Drongoski · Jun 9, 2009

The 2 curves intersect @ (0,0) and (1,1)

$\100dpi V\ \ =\ \ \pi \int_{0}^{1}\left [ (\sqrt{y})^{2}\ -\ (y^{2})^{2} \right ]\ dy \ \ =\ \ \pi \int_{0}^{1}(y-y^{4})\ dy\\ \\ \ =\ \ \pi \left [ \frac{y^{2}}{2}-\frac{y^{5}}{5} \right ]^{1}_{0}\ =\ \pi \left [ (\frac{1}{2}-\frac{1}{5})\ -\ 0 \right ]\ \ =\ \ \frac{3\pi }{10}\ \ unit^{3}$

Don't know if answer is right.

Aerath · Jun 9, 2009

I got 3pi/10 too. But just make sure you're using shell method [I'm not quite sure Drongoski used it].

Drongoski · Jun 9, 2009

Aerath said:
I got 3pi/10 too. But just make sure you're using shell method [I'm not quite sure Drongoski used it].

You are right. I forgot and was hurrying off to tutor.

Drongoski · Jun 9, 2009

Cylindrical Shells Version

$\100dpi V\ \ =\ \ \int_{0}^{1}2\pi x\ (y_{(2)}\ -\ y_{(1)})\ dx\ \ =\ \ 2\pi \ \int_{0}^{1}x(\sqrt{x}-x^{2})\ dx\\ \\ \ =\ \ 2\pi \int_{0}^{1}(x^{\frac{3}{2}}-x^{3})\ dx\ \ =\ \ 2\pi \left [ \frac{2}{5}x^{\frac{5}{2}}- \frac{x^{4}}{4} \right ]^{1}_{0}\ \ =\ \ 2\pi \left [ (\frac{2}{5}-\frac{1}{4}) -(0-0)\right ]\\ \\ =\ \ \ \frac{3\pi }{10}\ \ unit^{3}$

asb_92 · Jun 9, 2009

im confused..
could u please explain how u got to the answer?:S

haha im a losty when it comes to mathh ><
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Drongoski · Jun 9, 2009

asb_92 said:
im confused..
could u please explain how u got to the answer?:S

haha im a losty when it comes to mathh ><
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which part are u not clear about ?

asb_92 · Jun 9, 2009

the 2pi x( y2 - y1) part...:S
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Drongoski · Jun 9, 2009

Take a typical very thin vertical strip, of thickness delta x, of the area between the 2 curves; when spinned around the Y-axis, it forms a cylindrical shell, of radius x, and of height given by the 'y' of the upper curve minus the 'y' of the lower curve (y=x^2).

The circumference of a circle radius r is 2 x pi x r (for this question, it is "2 pi x" ); multiply this by the height (sqrt(x) - x^2) times delta x (like dx) gives you the volume of this thin cylinder. The total volume comes from adding up all these thin cylindrical shells, all the way from x=0 to x=1. This is roughly what the integral is all about.

Hope this helps.

asb_92 · Jun 9, 2009

http://community.boredofstudies.org/members/drongoski/
thanks heappsss!! that helped a lott!!! Drongoski

kwabon · Jun 10, 2009

Drongoski said:
The 2 curves intersect @ (0,0) and (1,1)

$\100dpi V\ \ =\ \ \pi \int_{0}^{1}\left [ (\sqrt{y})^{2}\ -\ (y^{2})^{2} \right ]\ dy \ \ =\ \ \pi \int_{0}^{1}(y-y^{4})\ dy\\ \\ \ =\ \ \pi \left [ \frac{y^{2}}{2}-\frac{y^{5}}{5} \right ]^{1}_{0}\ =\ \pi \left [ (\frac{1}{2}-\frac{1}{5})\ -\ 0 \right ]\ \ =\ \ \frac{3\pi }{10}\ \ unit^{3}$

Don't know if answer is right.

Aerath said:
I got 3pi/10 too. But just make sure you're using shell method [I'm not quite sure Drongoski used it].

he used the annulus/washer method. reckon annulus method would have been easier if the question didn't ask you what method to use???

Volume Question - shell method (1 Viewer)

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