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For the picture attached I am wondering why I cannot take the limits from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. If I take those limits, Sine vanishes (in the second last step) and the answer varies significantly. Where am I going wrong?

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    $\begingroup$ John ,thanks for edit .Can you please ,help me with the problem also? $\endgroup$ – shubham jain Aug 12 at 16:00
  • $\begingroup$ I believe it would really help if you sketched out the various solids on a coordinate plane. $\endgroup$ – JohnColtraneisJC Aug 12 at 16:07
  • $\begingroup$ Sorry for the bologna mate! $\endgroup$ – JohnColtraneisJC Aug 12 at 17:08

Integrating from $0$ to $\pi$ seems very strange to me too! Those bounds would correspond to the region $y \ge 0$, which doesn't match up with the fact that the cylinder in question lies entirely in the region $x \ge 0$ (if $a>0$), since its equation can be written as $(x-a/2)^2+y^2 \le (a/2)^2$.

So you're right, the integration should go from $-\pi/2$ to $\pi/2$, but you mustn't forget that when you substitute $r=a\cos \theta$ into $(a^2-r^2)^{3/2}$ you get $$ (a^2 - a^2 \cos^2 \theta)^{3/2} = (a^2)^{3/2} (\sin^2 \theta)^{3/2} = a^3 \sin^2 \theta \, \sqrt{\sin^2 \theta} = a^3 \sin^2 \theta \, | \sin \theta | . $$ Note the absolute value here, coming from the identity $\sqrt{t^2} = |t|$.

With this integrand, the integral from $-\pi/2$ to $0$ will give the same contribution as the integral from $0$ to $\pi/2$, and you don't get zero as your answer anymore.

  • $\begingroup$ I've been monitoring this problem and was actually struggling a bit my own self...thank you for coming to the rescue. +1! $\endgroup$ – JohnColtraneisJC Aug 12 at 17:07
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    $\begingroup$ Awesome Hans .Thanks a lot .A usual error of taking √x^2 as x and not modulus x made me pay the price .I was doubting my concepts and was in terrible frame of mind .Thanks for the timely rescue . $\endgroup$ – shubham jain Aug 12 at 17:32

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