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I have a fundamental doubt .Suppose , we are integrating a double integral in X-Y plane , with parametric equation reducing to $$ r= a \cos{t} $$.Then , I am of the view that limits of t should ideally vary from -π/2 to +π /2 .But , there are many text books ,taking the limits for t ( while doing the double integral ) as 0 to π .Is it justified .At π ,radius vector is negative ,so how can limit vary till π.Please ,if anyone can justify.

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  • $\begingroup$ It depends on what region are you integrating. The limits can't be decided just by the function. We need some more information. $\endgroup$ – HS Singh Aug 12 at 12:25
  • $\begingroup$ We are integrating in entire region of circle . $\endgroup$ – shubham jain Aug 12 at 12:26
  • $\begingroup$ Can t be ever equal to π ?.At π circle won't be there .Circle is there in 1st and 4th quadrant. $\endgroup$ – shubham jain Aug 12 at 12:28
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    $\begingroup$ Can you upload the exact question? $\endgroup$ – HS Singh Aug 12 at 12:31
  • $\begingroup$ Ok so question is calculate the volume bounded by Sphere x^2 +y^2 + z^2 =a^2 and Cylinder x^2 +y ^2 = 2ax .Please try to solve it in parametric form . $\endgroup$ – shubham jain Aug 12 at 12:40
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For your particular example, i.e. the volume enclosed by a sphere and a cylinder, the parametric integral should take the form

$$\int_{0}^{\pi}\sin\theta d\theta\int_{-\pi/2}^{\pi/2} d\phi\int_0^{f(\theta,\phi)}r^2dr$$

The range for either angle integration is $\pi$.

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  • $\begingroup$ Quanto ,please if you evaluate it completely .I am still having some doubts . $\endgroup$ – shubham jain Aug 12 at 13:20
  • $\begingroup$ The actual integration is quite involved due to overlapping between the two shapes. I'll work it out when I get a chance. $\endgroup$ – Quanto Aug 12 at 15:44
  • $\begingroup$ math.stackexchange.com/questions/3321156/… now i have added solution to that question .Please visit the link .I have explicitly mentioned my exact doubt there .Please see if you can help me out . $\endgroup$ – shubham jain Aug 12 at 15:48
  • $\begingroup$ Result in the link does not seem right, which ought to have $\pi$. $\endgroup$ – Quanto Aug 12 at 17:55
  • $\begingroup$ Its a printing error in last step .Anyways , satisfactory answer has been given by Hans.Thanks $\endgroup$ – shubham jain Aug 12 at 18:03

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