Question

Write a triple integral that gives the specified volume.
Under the sphere x^2+y^2+z^2=4 and above the region x^2+y^2≤4, 0≤x≤1, 0≤y≤2 in the xy-plane.

I'm having trouble figuring out the limits of integration for y and x. If someone could help me out with that I would really appreciate it

Answer 1

and your approach?

Answer 2

we agree that dz dy dx is a fair movement?

Answer 3

sorry, what do you mean by fair movement?

Answer 4

the limits of integration are determined by how we move across the space

if we determine the movement of dz first, then dy, then dx we have covered the total space

Answer 5

for example, if we move along dz: we move from z=0 to z=f(x,y) right?

Answer 6

dy, y moves from 0 to sqrt(4-x^2) ... not 2-x^2 :)

Answer 7

dx is just given, 0 to 1

Answer 8

yes you're right, now what I am confused about is why the limits for why aren't just 0 to 2. I thought dy is just given like x is.

Answer 9

y*

Answer 10

we have a condition that is binding us to the inside of a circle; given the y interval and x interval ... how are we free to move?

Answer 11

when x=0, y=0 to 2
when x=what .23, y=0 to whatever the circle arc bind us to
when x=n, y=0 to whatever the circle arc bind us to

when x=1, y = 0 to y = sqrt(3)

Answer 12

as we move along the x axis, y depends on x
sufficient to the condition: y = sqrt(4-x^2)

Answer 13

as we move along the plane z is dependant upon a point (x,y) such that it takes on values between 0 and the sphere above it

Answer 14

if we move y from 0 to 2, then x is limited by 2 functions since we have a distruption in teh process.