I Q about finding area with double/volume with triple integral

[AutumnWater]

So when finding the Area from a double integral; or Volume from a triple integral: If the curve/surface has a negative region: (for areas, under the x axis), (for volumes, below z = 0 where z is negative)

What circumstances allow the negative regions to be taken into account as positive when finding the area/volume without having to split the integral up and add a negative piece to it?

Does polar form specifically allow for this since r can not be negative? Or is a double integral for area, triple integral for volume sufficient for this regardless of the coordinate system used since they are the same dimensionality as the scalar in question? Or they all have to be split up in order to calculate the negative area/volume correctly regardless of which coordinate system we're using?

 

[Mark44]

So when finding the Area from a double integral; or Volume from a triple integral: If the curve/surface has a negative region: (for areas, under the x axis), (for volumes, below z = 0 where z is negative)

If you're calculating area with a double integral, the integrand will be 1. Similarly, if you're calculating the volume of a solid in a triple integral, the integrand will also be 1. The iterated integral $\int_{y = c}^d~\int_{x = a}^b f(x, y) dx~dy$ gives the volume bounded by the graph of z = f(x, y) above the rectangular region defined by the limits of integration. For "volume" I'm assuming that the graph of f does not dip below the x-y plane.

What circumstances allow the negative regions to be taken into account as positive when finding the area/volume without having to split the integral up and add a negative piece to it?

None that I can think of.

Does polar form specifically allow for this since r can not be negative?

r can be negative. For example, the point $(-2, \pi/4)$ lies in the third quadrant.

Or is a double integral for area, triple integral for volume sufficient for this regardless of the coordinate system used since they are the same dimensionality as the scalar in question? Or they all have to be split up in order to calculate the negative area/volume correctly regardless of which coordinate system we're using?

If you blithely integrate over an interval where the integrand function dips below the x-axis (with an integrand y = f(x)) or below the x-y plane (with an integrand of z = f(x, y)), the value of the integral will be less than the actual area/volume.

 

[AutumnWater]

If you're calculating area with a double integral, the integrand will be 1. Similarly, if you're calculating the volume of a solid in a triple integral, the integrand will also be 1. The iterated integral $\int_{y = c}^d~\int_{x = a}^b f(x, y) dx~dy$ gives the volume bounded by the graph of z = f(x, y) above the rectangular region defined by the limits of integration. For "volume" I'm assuming that the graph of f does not dip below the x-y plane.
If you blithely integrate over an interval where the integrand function dips below the x-axis (with an integrand y = f(x)) or below the x-y plane (with an integrand of z = f(x, y)), the value of the integral will be less than the actual area/volume.

Thanks, I realize this is the case for double integral volumes and non volume triple integrals.
But I was specifically thinking when the integrands are 1 for double integral areas and triple integral volumes with boundaries that are in the negative regions. Does this situation allow the (1 integrand) double integral area or triple integral volume to be equal to the actual integral value itself? Or these also need to be split up?

Edit: I think I get it somehow now, it shouldn't matter that the x/y boundaries have negative y values for a double integral finding the area since z = 1 is a flat plane on the z = 1 level curve, and the area covered by this curve is always positive regardless of whether the y boundaries have become negative or not because the integral is actually trying to find the volume of the slice with height 0 and in turn finds the area instead, the sign of this slice's value is based on the z axis not on y-axis right?

 

[Mark44]

Thanks, I realize this is the case for double integral volumes and non volume triple integrals.
But I was specifically thinking when the integrands are 1 for double integral areas and triple integral volumes with boundaries that are in the negative regions. Does this situation allow the (1 integrand) double integral area or triple integral volume to be equal to the actual integral value itself? Or these also need to be split up?

It doesn't matter if the boundaries of the interval (for single integral) or region (for double or triple integrals) include negative parts. It only matters whether the integrand becomes negative.

Edit: I think I get it somehow now, it shouldn't matter that the x/y boundaries have negative y values for a double integral finding the area since z = 1 is a flat plane on the z = 1 level curve, and the area covered by this curve is always positive regardless of whether the y boundaries have become negative or not because the integral is actually trying to find the volume of the slice with height 0 and in turn finds the area instead, the sign of this slice's value is based on the z axis not on y-axis right?

Close, but not quite. The iterated integral $\int_{y = c}^d~\int_{x = a}^b 1~dy~dx$ gives the area of the region defined by the limits of integration. It doesn't matter if some or all of the limits of integration are negative, as long as $c \le d$ and $a \le b$. The integrand, z = 1, determines a solid whose thickness is 1 (not a height of 0 as you said), so that the volume of this solid is numerically equal to the area of the region of integration. Only the units (area units vs. volume units) are different.

 

[AutumnWater]

Thanks, that clears it all up for me now.