That looks good !whynot314 said:Homework Statement
use a double integral to find the volume of the solid bounded by.
z=x^2+2y^2 and z=12-2x^2-y^2
I want to change variables using polar coordinates, I know its the top minus the bottom, and the intersection between the two is a circle radius 2.
The Attempt at a Solution
I want to make sure i have the correct set up
[itex]\int^{2\pi}_{0}[/itex] [itex]\int^{2}_{0}[/itex] (12-3r[itex]^{2}[/itex])rdrd[itex]\theta[/itex]
The formula for calculating volume using double integrals in polar coordinates is V = ∫∫f(r,θ) r dr dθ, where f(r,θ) is the function representing the shape or object and the limits of integration are based on the bounds of r and θ.
Yes, double integrals in polar coordinates can be used to find the volume of any shape as long as the shape can be represented by a function in polar coordinates.
The region of integration is determined by the bounds of r and θ, which are determined by the shape or object being studied. The bounds must be carefully chosen to ensure that the entire region is covered and no overlap occurs.
Some common mistakes to avoid when using double integrals in polar coordinates include not carefully choosing the bounds of integration, forgetting to include the r in the integrand, and not converting the function to polar coordinates if it is given in Cartesian coordinates.
Double integrals in polar coordinates can be applied in real-world situations such as calculating the volume of a cylindrical or spherical tank, determining the mass and center of mass of an object with varying density, and finding the volume of a rotating solid using the method of cylindrical shells.