The volume of the region bounded by $y^2+z^2=1\ \text{and} \ z^2+x^2=1$ should be found. Cylindrical conditions are perhaps the most appropriate in this case but the limits of the integral are what confuse me. If so then the bounds for z are like this, $$2\int_{0}^{2\pi}\int_{0}^{1}\int_{\sqrt{1-r^2\cos^2\theta}}^\sqrt{1-r^2\sin^2\theta} rdzdrd\theta$$ or is it the opposite way around or does it even make a difference.


Indeed, cause of the symmetric, the final triple integrals are as:

$$V=8\left(\int_{\theta=0}^{\pi/4}\int_{r=0}^{\sec(\theta)}\int_{z=0}^{\sqrt{1-r^2\sin^2(\theta)}}+\int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{\csc(\theta)}\int_{z=0}^{\sqrt{1-r^2\cos^2(\theta)}}\right)$$ But I don't think this way of using the Cylindrical Coordinates make the integrals above easy to solve. I made the following plot. Note that the first integrals is got on the $I$ and another is for $II$:

enter image description here

I, personally, prefer the method which described in here

  • $\begingroup$ Can you please explain where I went wrong? $\endgroup$ – John Dec 12 '13 at 10:29
  • $\begingroup$ I don't see why the integrals had to be divided into? $\endgroup$ – John Dec 12 '13 at 10:35
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    $\begingroup$ Thank you for the perfect and thorough explanation and drawing. I understand now. I was wondering what if we changed the cylindrical coordinates to x being the constant and r,$\theta$ in y-z plane $\endgroup$ – John Dec 12 '13 at 11:07
  • $\begingroup$ @B.S.: Wow - excellent! +1 $\endgroup$ – Amzoti Dec 12 '13 at 14:02

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