The volume to be integrated is bounded by the planes
0≤x≤2
0≤y≤2
z=0
and the surface
z=16-x²-2y².
Thus the volume is given by the triple integral:
V=∫∫∫z dz dx dy
Where z is evaluated from 0 to 16-x²-2y²,
x and y are both evaluated from 0 to 2.
Thus
V=∫∫(16-x²-2y²)dx dy
=∫(88-12y²)/3 dy
=48
What is the volume of the structure bounded by the regions x^2+2y^2+z≤16, 0≤x≤2, 0≤y≤2 and z≥0?
1 answer
1 answer