myost
On the exam review:
What would an answer to number 6 look like? the question is:
Let R denote the top half of a sphere of radius 2. Set up the triple integral of the arbitrary function f in spherical coordinates.
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Exam review
knguyen3
I got this.
\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{2}f(\rho sin(\phi)cos(\theta), \rho sin(\phi)sin(\theta), \rho cos(\rho)) \rho^2 sin(\phi)d\rho d\phi d\theta.
\rho is radius from 0 to 2, \phi is the angle between the radius and z-axis and rotate from 0 to \pi/2 because only top half of the sphere, and \theta from 0 to 2\pi (full rotation about z-axis).
Maui
On the Exam review:
How do you deal with the “z” before rdrdθ after setting up the triple intergral? This is for problem 5.
Let R denote the region between f(x, y) = 9 − (x^(2) + y^(2)).
knguyen3
I think problem 5 is missing information, It should be bounded by the function f(x,y) and the xy-plane. Otherwise, I have no idea how to deal with ‘z’, it has no bound on the bottom and it could be \int_{-\infty}^{f(x,y)}dz but I think this isn’t right.
If the problem include bounded by xy-plane, it is similar with problem 3. The answer would be \int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{9-r^2}(r^2z)rdzdrd\theta