No, the radical is NOT of that form in polar coordinates. But you can modify the coordinate system to "elliptic coordinates". Let [itex]x= r a cos(\theta)[/itex], [itex]y= r a sin(\theta)[/itex]. Then [itex]x^2/a^2= r^2 cos^2(\theta)[/itex] and [itex]y^2/b^2= r^2 sin^2(\theta)[/itex]. You will need to use the Jacobian to get the correct differential dx dy is not just "[itex]r dr d\theta[/itex]" now.manenbu said:Homework Statement
Solve:
[tex]\iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1} \sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} dx dy [/tex]
Homework Equations
Cartesian to Polar
The Attempt at a Solution
Well - this Integral should be solved as a polar function (the radical should be [tex]\sqrt{1-r^2}ab[/tex] when expressed in polar coordinates. I just can't get this right. Guidance please?
Cartesian coordinates use the x- and y-axes to locate a point in a two-dimensional plane, while Polar coordinates use the distance from the origin (r) and the angle (θ) to locate a point.
Converting from Cartesian to Polar coordinates can make it easier to calculate the limits of integration and the integrand in a double integral, especially for integrals with circular or radial symmetry.
To convert a double integral from Cartesian to Polar coordinates, you need to replace the variables x and y with rcos(θ) and rsin(θ), respectively, and change the limits of integration to correspond to the new variables.
Cartesian coordinates are usually more suitable for rectangular or linearly symmetric regions, while Polar coordinates are better for circular or radial symmetric regions. Therefore, the choice between the two depends on the shape of the region being integrated.
Yes, using Polar coordinates in a double integral can become more complicated when the region of integration is not circular or radial symmetric. It may also require additional steps to convert the integrand into the new coordinate system.