integration (1 Viewer)

fullonoob · Mar 5, 2010

The shaded region is bound by the curve y= x^3, the y axis and the line y=8
Determine the volume of revolution formed when this region is rotated about the x axis. Working out and not procedure (instructions) please. Thx
Is the answer 576 pi /7? Confirm please

Tofuu · Mar 5, 2010

Pwnage101 · Mar 5, 2010

$Required\;volume\; =\pi \times 8^{2}\times 2-\pi \int_{0}^{2}\left (y \right )^{2}dx =128\pi -\pi \int_{0}^{2}\left (x^{3} \right )^{2}dx =128\pi -\pi \int_{0}^{2}x^{6}dx =128\pi -\frac{\pi}{7}\left [x^{7} \right ]_{0}^{2} =128\pi -\frac{\pi}{7}\left ( 2^{7}-0^{7} \right ) =128\pi -\frac{\pi}{7}\times 128 =\frac{768}{7}\pi \; units^{3}$

Remember what you are after is NOT the volume when the shaded area between y=x^3, x=2 and the x-axis is rotated about the x-axis, but the volume of the cylynder when you rotate the shaded region between y=8, x=2, and the x and y axes about the x-axis MINUS this.

Diagram would be useful.

hscishard · Mar 5, 2010

Pwnage101 said:
$Required\;volume\; =\pi \times 8^{2}\times 2-\pi \int_{0}^{2}\left (y \right )^{2}dx =128\pi -\pi \int_{0}^{2}\left (x^{3} \right )^{2}dx =128\pi -\pi \int_{0}^{2}x^{6}dx =128\pi -\frac{\pi}{7}\left [x^{7} \right ]_{0}^{2} =128\pi -\frac{\pi}{7}\left ( 2^{7}-0^{7} \right ) =128\pi -\frac{\pi}{7}\times 128 =\frac{768}{7}\pi \; units^{3}$

Remember what you are after is NOT the volume when the shaded area between y=x^3, x=2 and the x-axis is rotated about the x-axis, but the volume of the cylynder when you rotate the shaded region between y=8, x=2, and the x and y axes about the x-axis MINUS this.

Diagram would be useful.

Give that man the stamp of approval :uhhuh:. I finally understand the stupid forumla.

fullonoob · Mar 5, 2010

Tofuu said:
$y=x^3$
$y^2=x^6$
$wheny=8\rightarrow x=2$
$v=\pi\left [ \frac{x^7}{7} \right ]$ between x=0 and x=2
$v=\frac{128\pi}{7}units^3$

this is what i got

this is incorrect as you are not finding the area bound by y=8 as well

fullonoob · Mar 5, 2010

Pwnage101 said:
$Required\;volume\; =\pi \times 8^{2}\times 2-\pi \int_{0}^{2}\left (y \right )^{2}dx =128\pi -\pi \int_{0}^{2}\left (x^{3} \right )^{2}dx =128\pi -\pi \int_{0}^{2}x^{6}dx =128\pi -\frac{\pi}{7}\left [x^{7} \right ]_{0}^{2} =128\pi -\frac{\pi}{7}\left ( 2^{7}-0^{7} \right ) =128\pi -\frac{\pi}{7}\times 128 =\frac{768}{7}\pi \; units^{3}$

Remember what you are after is NOT the volume when the shaded area between y=x^3, x=2 and the x-axis is rotated about the x-axis, but the volume of the cylynder when you rotate the shaded region between y=8, x=2, and the x and y axes about the x-axis MINUS this.

Diagram would be useful.

sorry but also incorrect. you can't assume it forms a circle, as it has diff radii. it does not form a cylinder. oh yeah find the X>0 part not <0, forgot to add that in.
The shaded region is between y = 8 and x^3 (x>0 part of graph)
I did it differently but not sure if its right so yeah

Pwnage101 · Mar 5, 2010

I did not find the volume of the cylender as the explicit answer;

would you agree that:

The volume when the region bounded by the curve y=x^3, x=2 and the x-axis (x>0) is rotated about the x-axis

PLUS

The volume when the region bounded by the curve y= x^3, the y axis and the line y=8
(x>0) is rotated about the x-axis [Which is our Required Volume]

EQUALS

The volume when the region bounded by the lines y=8, x=2, and the x and y axes is rotated about the x-axis [Which is a cylinder, radius 8, height 2].

Rearrange to get the required area: i.e.

The volume when the region bounded by the lines y=8, x=2, and the x and y axes is rotated about the x-axis [Which is a cylinder, radius 8, height 2].

MINUS

The volume when the region bounded by the curve y=x^3, x=2 and the x-axis (x>0) is rotated about the x-axis

EQUALS

The volume when the region bounded by the curve y= x^3, the y axis and the line y=8
(x>0) is rotated about the x-axis [Which is our Required Volume]

hscishard · Mar 5, 2010

fullonoob said:
sorry but also incorrect. you can't assume it forms a circle, as it has diff radii. it does not form a cylinder. oh yeah find the X>0 part not <0, forgot to add that in.
The shaded region is between y = 8 and x^3 (x>0 part of graph)
I did it differently but not sure if its right so yeah

...? They do dont they? Same radius I mean. Don't know where you got circle from.

fullonoob · Mar 5, 2010

ahhh i see
the answer is effed then :L
thx

hscishard · Mar 6, 2010

?? His answer and steps made perfect sense. Or did you mean your answers...meh. Man gl to us in the hsc

mdmm92 · Mar 27, 2010

integration (1 Viewer)

fullonoob

fail engrish? unpossible!

Tofuu

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Pwnage101

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hscishard

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fullonoob

fail engrish? unpossible!

fullonoob

fail engrish? unpossible!

Pwnage101

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hscishard

Active Member

fullonoob

fail engrish? unpossible!

hscishard

Active Member

mdmm92

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