15.8.4: Setting Up an Integral That Gives the Volume Inside a Sphere and Below a Half-Cone

Here's another way to get the lower bound on phi, assuming z=sqrt(x^2+y^2). Set y=0 to get z=sqrt(x^2) = x (if you take the non-negative root), which means phi = pi/4, since z = x is the 45-degree line in the 1st quadrant of the xz-plane.

If that's not easy to see, then draw the corresponding triangle (as in the video) and get tan(phi)=1/1 =1. Inverse tan gives phi = pi/4.

[reidflemingworldstoughestm]

Finding phi1:
z=sqrt (x^2+y^2).
When y=0 the equation becomes z=x, with slope 1. This slope has the angle from the vertical of pi/4, so
phi1=pi/4.

For z= sqrt (3x^2+3y^2).
y=0 -> z=sqrt(3)x, slope=sqrt(3).
Slope is tan(angle from horizontal) and slope is cot(angle from vertical), so
phi1 = arctan(1/sqrt(3)) = pi/6
or
phi = pi/2-arctan(sqrt(3)) = pi/6.

The angle of the side of the cone, phi, is quite literally sitting right there in the coefficient in front of the sqrt(x^2+y^2). Just invert the coefficient and take the arctan.

[sbusisomcebongwenya]

Well. Im saving this one. Thank you sir!!!

[sandlertossone]

Won't a cone always hit the sphere at pi/4?

[bangaloremathematicalinsti]

what is the software you are using to draw the graphs?

[MrSqueakinator]

Don't you need to be integrating z for your first integral?

[KishoreG]

I dont understand. If you rewrite the cone equation with z = pcos(phi), x = psin(phi)cos(theta), and y = psin(phi)sin(theta), you will get sin(phi) = cos(phi), which implies phi = pi/4, not pi/6. Could you please explain?