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The cone has the formula: $x^2 + y^2 = z^2 , 0≤z≤2$ So I used the cylindrical coordinates to get the following answer: $$\int_0^{2\pi}\int_0^2\int_0^2 dz\,rdr\,d\theta = 8\pi$$

In the solution of the doctor, he used spherical coordinates as follows: $$\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sec\Phi}\rho^2\sin\Phi \,d\rho \,d\Phi \,d\theta=\frac{8\pi}{3}$$

  1. Why is my answer wrong?
  2. Why, in doctor's solution, $\rho=2\sec\Phi$ in the upper limit of the integration?
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  1. Your answer gives you the volume of a cylinder with height $2$.

  2. See the picture:

enter image description here

$$\frac{2}{\rho}=\cos \phi\implies \rho=\frac{2}{\cos\phi}=2\sec\phi$$

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  • $\begingroup$ Thank you. But for (1), I know that it gives the volume of that cylinder but I want to know what is my wrong in setting integral limits? Can't I get the volume of the cone using cylindrical coordinates (for z, we always enter the region from z=0 and exit from z=2. For r, the shadow of the cone is a circle on the xy-plane with r=2 so r ranges from 0 to 2) ? Isn't it done like this? $\endgroup$ – ammar May 30 '15 at 11:46
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    $\begingroup$ @ammarx: If you use cylindrical coordinate, remember your $z$ starts from the boundary of the cone to $2$. It is not from $0$. So you need to express your lower bound of $z$ using $r,\theta$ as in the spherical coordinates. $\endgroup$ – KittyL May 30 '15 at 12:56

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