In our initially method of solving this problem, we decided to break the region R into three sub-regions. Our final answer resulted in calculating the double integral of the function  for each of the sub-regions and then summing them, that is

While this method indeed works, we ended up with some complicating numerical values for our answer. Let’s see whether there is a less tedious solution in terms of evaluating the integrals.

A quick look at the region R and we see that we can divide this region in a different way. We know that the double integral gives us a signed volume. So just like how we summed the individual sub-regions, why not we try subtracting sub-regions from say a bigger sub-region which ultimately gives us back the region R.

From the diagram, we can say

where  is the sub-region bounded by  and  and  is that bounded by  and . So the double integral of region R  can now be expressed as

Is this fine? Yes it is simply because we are adding or subtracting signed volumes accordingly to get our desired region. Moreover, we are dealing with just two double integrals now, one less from before.

Moreover, notice that both sub-regions are bound by exactly two  equations. There is no issue of intersecting points in these regions. So proceeding with the calculations,

Hmmm, that’s a little peculiar. We already got our answer without subtracting the answer from . We may be quick to find an error in the calculation, but there’s isn’t. Evaluating the double integral for ,

and thus

,

which is the same result as before. So we may now wonder why the double integral evaluated over the region  is 0. Well, there is an easy explanation.

The double integral gives us a signed volume. Now obviously, we know that in the sub-region  as bounded by  and , the function (height function in this case) cannot be , but when given as , we either get a positive or negative value for  in this region. So, the positive signed volume and negative signed volume in  will perfectly balance out when we combine them together resulted in 0 volume.

Be mindful that this does not mean no volume. It means that the volume above and below the xy-plane are equal. We can also say that the function  will cut the plane in this region , as graphically shown below.

So there you have it. An alternative way for our hunting for gold problem. I emphasize again that the negative volume suggest to us a cave in the designated location and so cranes are needed to dig for the gold.

Then again, you could just focus on the mathematics and leave the gold digging to others.