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When finding the volume of rotation, along x axis doesn't give an answer

Posted 9 years ago

for example, https://www.wolframalpha.com/input/?i=volume+of+revolved+solid+of+region+bounded+by+y+%3D+-2x+%2B+1+%2C+y+%3D+5%2C+x+%3D+0+about+the+x+axis

doesn't give a definite value of the answer to the question, why is that?

On the other hand, the same question, except rotating along y axis, gives an answer: https://www.wolframalpha.com/input/?i=volume+of+revolved+solid+of+region+bounded+by+y+%3D+-2x+%2B+1+%2C+y+%3D+5%2C+x+%3D+0+about+the+y+axis

This is creating difficulties, and any help is deeply appreciated.

POSTED BY: Muhibur Rahim

Neither the first nor the second input give a value to me, but

In[41]:= (* x Rotation *)
\[Pi] Integrate[(-2 x + 1)^2, {x, -2, 0}]
Out[41]= (62 \[Pi])/3

In[42]:= (* y Rotation *)
\[Pi] Integrate[(-1/2 (y - 1))^2, {y, 0, 1}] + \[Pi] Integrate[(1/2 (y - 1))^2, {y, 1, 5}]
Out[42]= (65 \[Pi])/12

because in cylindrical co-ordinates the volume is $\int_a^b\int_0^{2\pi}\int_0^{f(z)} r dr d\varphi dz = 2\pi \int_a^b \frac{1}{2} (f(z))^2 dz = \pi \int_a^b (f(z))^2dz$ where $f(z)$ is understood to be non-negative, i.e. if $f(z)$ has a zero one has to stop integration at it and continue with the next positive part of the boundary description becoming $f(z)$ and so on.

POSTED BY: Udo Krause
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