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

[@reidflemingworldstoughestm1394]

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.

[@cchamberlin2000]

True. I mention that in reply to one of the older questions.

[@sbusisomcebongwenya167]

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

[@rashedbinkhonain7022]

Great video!

[@Srijjjj]

Good one sir

[@sandlertossone1813]

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

[@calculus3d130]

No. Use z=sqrt(3x^2+3y^2) for your half-cone. You should get phi=pi/6

[@calculus3d130]

Here's another way to get phi, assuming z=sqrt(3x^2+3y^2). Set y=0, which turns the side of the cone into a line in the xz-plane. You'll get z=sqrt(3)*x. Draw the corresponding triangle and get tan(phi)=1/sqrt(3) or phi=pi/6.

[@nathannixon2727]

What if the cone's summit isint centered with the sphere?

[@calculus3d130]

@@nathannixon2727 If the cone's vertex doesn't coincide with the center of the sphere, then it is potentially a harder problem for which spherical coordinates may not be helpful. Still, in general, to find the volume of a solid region, you would triple integrate: the inner lower limit of integration would be the function that gives the lower boundary surface, while the inner upper limit would be the function that represents the upper boundary surface.

[@bangaloremathematicalinsti5351]

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

[@calculus3d130]

OneNote and a cheap Wacom drawing tablet.

[@lol-gu4wr]

nice video

[@MrSqueakinator]

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

[@calculus3d130]

You could do it that way in rectangular or cylindrical coordinates. I used spherical coordinates here, so no need to worry about z.

[@KishoreG2396]

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?

[@calculus3d130]

In the example from the video, the lower value of phi does equal pi/4. If you're referring to the example z=sqrt(3x^2+3y^2), then, no, phi must be smaller than pi/4 because this cone is "taller" than the one from the video, given by z=sqrt(x^2+y^2). To see this, set y=0 in z=sqrt(3x^2+3y^2). This turns the side of the cone into a line in the xz-plane. You'll get z=sqrt(3)*x. Draw the corresponding triangle and get tan(phi)=1/sqrt(3) or phi=pi/6. Remember, the lower limit on phi in this case is measured from the z-axis to the side of the cone (or line if you look at the two-dimensional cross-section you get by setting y=0).

[@KishoreG2396]

@@calculus3d130 I understand now. Thank you.

[@calculus3d130]

@@KishoreG2396 Glad to help.

[@jumanamohamediqbal6531]

Isn't it 0 to π/4

[@calculus3d130]

No, assuming you're referring to the limits on phi, that would give the volume inside the cone and the sphere.