y = x
y = sqrt(3)*x
y = x2 + y2
There is a unique region bounded by al three curves; that is what we are calling R. The problem that we'll consider here is the computation of the area of R, and the volume swept out by rotating R arount the y-axis.
This problem is most easily done in polar coordinates. But it can also be done in Cartesian. We will do both here because both are instructive, as is the comparisson that doing the problem both ways permits.
First, lets look at the problem in the Cartesian Integration Applet. This will graph the region for us, and estimate the integral. This provides a check on our work. We enter the functions into the applet as follows:
Notice the colored squares; the curve coming from each function will be drawn in the color of the corresponding square so you can tell what is what. Also, f(x,y) is the integrand which is just 1 here for finding the area.
Entering all of these in, and hitting return in each text field when its got what you want, we get the graph:
In this graph, the center is at (0,0), and the width and height are both 1.2. The region in question occupies a relatively small fraction of the graph, so many of the sqaure that overlap the region do so on the edge -- there are 70 boundary sqaures and only 40 complete squares in the interior.
The lower quadrature sum is what you get by taking the squares completely inside the region, and adding up their areas.
This gives us lower bound on thee number we are trying to compute; that is, the exact answer is a number bigger than this.
But we also get an upper bound: If we add up the area of all the sqaures that touch the region in any way, we get the upper quadrature sum.
The exact answer lies somewhere between the two. The applet also reports the average of these as a good guess for the true value.
The applet can be used to check our work: If we get an answer that is outside the range of values between the lower and upper quadrature sums, then a mistake was made.
Still, this range is very big right now. Can we shrink it? The answer is yes. We just need to zoom in on the region. In the next picture, we have typed new values in the graph range panel. The center is now at (0.25,0.4), and the width and height are 0.8.
Things are much better -- there are now more interior squares than boundary sqaures. For this reason, there is a less significant difference when one includes or leaves out the boundary sqaures, and the upper and lower quadrature sums are much closer together.
Still the interval is pretty wide. One has to use a very, very large number of squares to close the gap down to a really small number.
But you can also see that about half of the
light blue stuff is in the region and half out.
So we expect the average of the
upper and lower quadrature sums to be pretty close
Solution in Cartesian Coordinates