Limits of integration for parametric equation

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?

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

1 Answer

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.

• I've been monitoring this problem and was actually struggling a bit my own self...thank you for coming to the rescue. +1! – JohnColtraneisJC Aug 12 at 17:07
• 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 . – shubham jain Aug 12 at 17:32