Question

by direct integration find the area of that portion of the sphere x^2+y^2+z^2=25 which falls between planes z=3 and z=4

Answer 1

the region D that we will be integrating over is\[x^2+y^2+3^2=25\implies x^2+y^2=16\]

Answer 2

I suppose we can put this into cylindrical coordinates

Answer 3

how?

Answer 4

\[\iiint dV=\iint\limits_D f(x,y)dA=\iint\limits_D 25-x^2-y^2dA\]the region as I mentioned is a circle of radius 4\[D=\{(r,\theta)|0\le r\le 4;0\le\theta\le2\pi\}\]so we get\[\iint\limits_D 25-r^2dA=\int_0^{2\pi}\int_0^4 25r-r^3drd\theta\]

Answer 5

no radius is between 4 and 3

Answer 6

no, I put the region we are integrating over into polar coordinates

Answer 7

the radius is a constant of taken to be from the origin, because this is a sphere

Answer 8

if taken*

Answer 9

the question hinges on what coordinate system we are using

Answer 10

yes, when we changed to polar coordinate, then the radius will from 4 to 3 because for z=3 it will give us radius of 4 and for z=4 then it gives radius of 3

Answer 11

I rewrote z as a function of x and y and put the region of integration into polar coordinates

z is a function of x and y (i.e. where we are in the x,y plane) so son't confuse that with radius, which we only used for the region

Answer 12

okay

Answer 13

the region is the whole circle because we have to chose a shape that lies *below* the thing we are integrating (like the shadow that would be made by shining a light on the object toward the xy-plane)

Answer 14

http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx

Answer 15

thanks

Answer 16

I was also able to simplify the integral by plugging\[x=r\cos\theta\]\[y=r\sin\theta\]so as you can see radius is again a function of x and y

Answer 17

welcome