spin solids. help integral

**berto** · March 28th 2011, 07:36 PM

hello. i have problem with spin solid to find volume.

this is google translate of problem with help of math writing online program:

find the volume of the area bounded by: $x = 1-y^4$ , x = 0 around x=2

$x = 1-y^4$ is same as $y=\sqrt[4]{1-x}$ , yes? so is ring method:

$\pi\int_{0}^{1}(x-1)^2 -(\sqrt[4]{1-x})^2 dx$

this is okay?

thank you. sorry for bad english

**TKHunny** · March 28th 2011, 08:36 PM

First, $x = 1 - y^{4}$ MAY be the same as $y = \sqrt[4]{1-x}$ , depending on what you are doing. Be more careful.

Second, my rule, ALWAYS do it both ways so that you gain experience and you can check your own answer.

Washers: $\int_{-1}^{1}\pi\cdot\left[2^{2}-(2-x)^{2}\right]\;dy$ Substitute for 'x' and evaluate. With consideration for symmetry, this could be $2\cdot\int_{0}^{1}\pi\cdot\left[2^{2}-(2-x)^{2}\right]\;dy$

Cyllinders: $\int_{0}^{1}2\cdot\pi\cdot (2-x)\cdot\left[2\cdot y\right]\;dx$ . Substitute for 'y' and evaluate. The 2y is there ONLY because of symmetry, otherwise, it would have been uglier.

**berto** · March 29th 2011, 07:05 AM

sorry. i dont get.

where is original function in integral? original function not needed?

graph shows half circle from -1 to 1, yes? if axis of spin is x=2, then we know radius of both: big radius = 3, little radius = 1

$\pi\int_{-1}^{1}3^2 - 1^2 dx$

$\pi\int_{-1}^{1}8dx$

$8\pi{x}\mid_{-1}^{1} = 8\pi - (-8\pi) = 16\pi$

is this answer?

i think i answer my question, yes? jajaja

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