Dhamnekar Win,odd Posted June 1, 2022 Share Posted June 1, 2022 Find the volume V inside both the sphere x2 + y2 + z2 = 1 and the cone z = [math]\sqrt{x^2+ y^2}[/math] How to use change of variables technique in this problem? Link to comment Share on other sites More sharing options...
joigus Posted June 1, 2022 Share Posted June 1, 2022 Is this homework? Link to comment Share on other sites More sharing options...
NTuft Posted June 1, 2022 Share Posted June 1, 2022 Itchy trigger finger: I want to integrate by parts... Link to comment Share on other sites More sharing options...
mathematic Posted June 1, 2022 Share Posted June 1, 2022 Change (x,y) to polar coordinates. 1 Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted June 2, 2022 Author Share Posted June 2, 2022 9 hours ago, mathematic said: Change (x,y) to polar coordinates. My attempt: I graphed the cone inside the sphere as in my first post. But I don't understand how to use the change of variables technique here to find the required volume. My answer without using integrals is volume of the cone + volume of the spherical cap = Is this answer correct? If correct, how to derive this answer using integration technique? Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted June 3, 2022 Author Share Posted June 3, 2022 On 6/2/2022 at 1:09 PM, Dhamnekar Win,odd said: My attempt: I graphed the cone inside the sphere as in my first post. But I don't understand how to use the change of variables technique here to find the required volume. My answer without using integrals is volume of the cone + volume of the spherical cap = Is this answer correct? If correct, how to derive this answer using integration technique? I want to correct the volume asked in the question computed by me = [math] \frac{\pi}{3} \times \frac12 \times \frac{1}{\sqrt{2}} + \frac{\pi}{3} \times (1-\frac{1}{\sqrt{2}})^2 \times (\frac{3}{\sqrt{2}} -(1-\frac{1}{\sqrt{2}}))=0.534497630798[/math] Link to comment Share on other sites More sharing options...
Markus Hanke Posted June 3, 2022 Share Posted June 3, 2022 On 6/2/2022 at 2:39 PM, Dhamnekar Win,odd said: But I don't understand how to use the change of variables technique here to find the required volume. 1. Write down the transformation functions x=..., y=..., z=... for spherical coordinates 2. Calculate the Jacobian matrix from these 3. Calculate the determinant of the Jacobian matrix, in order to get dV=... 4. Determine the integration limits in your new coordinates - radius and equatorial angle are trivial, but for the polar angle you need to find the intersection of the ball and the cone (hint: eliminate z from the equations in the OP) 5. Write down the volume integral using your limits and volume form, and evaluate 1 Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted June 3, 2022 Author Share Posted June 3, 2022 (edited) 5 hours ago, Markus Hanke said: 1. Write down the transformation functions x=..., y=..., z=... for spherical coordinates 2. Calculate the Jacobian matrix from these 3. Calculate the determinant of the Jacobian matrix, in order to get dV=... 4. Determine the integration limits in your new coordinates - radius and equatorial angle are trivial, but for the polar angle you need to find the intersection of the ball and the cone (hint: eliminate z from the equations in the OP) 5. Write down the volume integral using your limits and volume form, and evaluate I computed the volume inside both the sphere [math] x^2 + y^2 + z^2 =1[/math] and cone [math] z= \sqrt{x^2 + y^2}[/math] as follows: [math]\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^1 \rho^2 \sin{\phi}d\rho d\phi d\theta= 0.61343412301 =\frac{(2-\sqrt{2})\pi}{3}[/math]. This answer is correct. Edited June 3, 2022 by Dhamnekar Win,odd 1 Link to comment Share on other sites More sharing options...
Markus Hanke Posted June 4, 2022 Share Posted June 4, 2022 Looks correct to me, this is what I got too 👍 Link to comment Share on other sites More sharing options...
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