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anonymous
 one year ago
Please help me with this calculus question!
anonymous
 one year ago
Please help me with this calculus question!

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phi
 one year ago
Best ResponseYou've already chosen the best response.4the first equation is a sphere centered at (0,0,1/2) with radius 1/2. to see this, complete the square on the z variable \[ x^2 + y^2 + z^2 =z \\ x^2 + y^2 + z^2z +(1/4) = 1/4 \\ x^2+ y^2 + (z\frac{1}{2})^2 = \frac{1}{4} \] in cylindrical coords, \[ r^2 + \left(z\frac{1}{2}\right)^2 = \frac{1}{4} \\ r= \sqrt{ \frac{1}{4} \left(z\frac{1}{2}\right)^2 } \]

phi
 one year ago
Best ResponseYou've already chosen the best response.4the second equation is a cone with its apex at the origin \[ z= \frac{1}{\sqrt{3} }\sqrt{x^2+y^2} \] in cylindrical coords \[ z = \frac{r}{\sqrt{3} } \\ r = \sqrt{3} \ z \]

phi
 one year ago
Best ResponseYou've already chosen the best response.4the composite shape is similar to an icecream cone: dw:1438214470899:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.4the cone intersects the sphere at a particular z. fix y=0. the second equation gives us \(x = \sqrt{3} z \) plug this (and y=0) in the first equation to get \[ x^2 + y^2 + z^2 =z \\ 3z^2 + z^2 = z\\ 4z^2  z = 0 \] we find z=0 (i.e the cone's apex intersects the sphere at the origin) and 4z1=0 or \( z= \frac{1}{4} \) which is what we want.

phi
 one year ago
Best ResponseYou've already chosen the best response.4we can now find the volume of the bottom cone, from z=0 to 1/4 in cyl coords the area element is r dr d\(\theta\) and the iterated integral is \[ \int_0^\frac{1}{4} \int_0^{2\pi} \int_0^{\sqrt{3}\ z} r \ dr\ d\theta\ dz \] visually we are doing dw:1438215237071:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.4above the cone, we integrate volume of the sphere \[ \int_{\frac{1}{4}}^1 \int_0^{2\pi} \int_0^{\sqrt{ \frac{1}{4} \left(z\frac{1}{2}\right)^2 }} r \ dr\ d\theta\ dz \] visually dw:1438215437879:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.4the sum of those two integrals will be 5pi/32
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