If we were to look straight down the \(z\) axis onto the region we would see a circle of radius 4 centered at the origin. This is because the top of the region, where the elliptic paraboloid intersects the plane, is the widest part of the region. We know the \(z\) coordinate at the intersection so, setting \(z 16\) in the equation of the paraboloid gives, \ which is the equation of real cars for kids a circle of radius 4 centered at the origin. Read the Report Cloud Security Threat Report (CSTR) Adapting to the new reality of evolving cloud threats. Here are the inequalities for the region and the function we'll be integrating in terms of polar coordinates.
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For instance, let's suppose we wanted to do the following integral, \ To this we would have to determine a set of inequalities for \(x\) and \(y\) that describe this region. These would be, \ With these limits the integral would become, \ Due to the limits on the inner integral this is liable to be an unpleasant integral to compute. However, a disk of radius 2 can be defined in polar coordinates by the following inequalities, \ These are very simple limits and, in fact, are constant limits of integration which waluigi plush almost always makes integrals somewhat easier. So, if we could convert our double integral formula into one involving polar coordinates we would be in pretty good shape. The problem is that we can't just convert the \(dx\) and the \(dy\) into a \(dr\) and a \(d\theta \).