Consider the solid that is bounded below by the cone z = sqrt{3x^{2}+3y^{2}} and above by the sphere x^{2} +y^{2} + z^{2} = 16..Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.

babeeb0oL 2021-03-05 Answered
Consider the solid that is bounded below by the cone z=3x2+3y2
and above by the sphere x2+y2+z2=16..Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.
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Expert Answer

averes8
Answered 2021-03-06 Author has 92 answers

Step 1
To set the triple integral in cylindrical coordinates
By using relation,
x=rcosθ
y=rsinθ
z=z
Thus,
The cone z=3x2+3y2 in cylindrical coordinate becomes,
z=3r2=3r
And the sphere become r2+z2=16
To find the limit of r,
Consider,
3r2=16r2
3r2=16r2

r2=4
r=2
Step 2
Thus, we can describe the region as
E={(r,θ,z)|0θ2π|0r23rz16r2}
Hence, the triple integral for the volume by cylindrical coordinates is
V=02π023r16r2rdzdrdθ
Step 3
Now, to set the triple integral in spherical coordinates
Since,
p2=x2+y2+z2
tanθ=yx
φ=arccos(zx2+y2+z2)
The sphere x2+y2+z2=16 gives p2=16=>p=4
From the cone
z=3x2+3y2=3r
pcos(φ)=3psin(φ)
ptan(φ)=13
φ=π6
Step 4
Thus, we can describe the region as
E={(p,φ,θ)|0p4,0φπ6,0θ2π}
Hence, the triple integral for the volume of the solid by spherical coordinate is
V=02π0π604p2sin(φ) d pd φ d θ
Step 5
Now, evaluating the integral of cylindrical coordinate we get.
V=02π023r16r2rdzdrdθ
V=17.9582
And evaluating the integral of spherical coordinate
V=02π0π604r2sin(φ)dpdφdθ
V=17.9582
Thus, by both coordinate systems, we get the same volume.
Therefore, both triple integrals are appropriate.

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