A region in the \(xy\)-plane is called type I if it is defined as the region between two functions of \(x\):

A region in the \(xy\)-plane is called type II if it is defined as the region between two functions of \(y\):

Let \(R\) be the type I region lying between \(y=g(x)\) and \(y=h(x)\) for \(a\le x\le b\). Then

\[\iint\limits_R f(x,y)\, dA = \int_a^b\int_{g(x)}^{h(x)} f(x,y)\, dy\, dx.\]

To visualize this, imagine the slice at a fixed value of \(x\). This slice begins at \(g(x)\) and ends at \(h(x)\). Shown below-left is the region in the \(xy\)-plane, with the slice drawn as a blue arrow, and below-right is the slice drawn in a 3D-coordinate system. (Note that the origin in the 3D-coordinate system is at the bottom right of the box.)

The area under this slice is \[\text{Area}(x)=\int_{g(x)}^{h(x)} f(x,y)\, dy.\] As with rectangular regions, then, to compute the original double integral, we sweep through the entire volume as these slices range from \(x=a\) to \(x=b\):

This yields the final result of \[\iint\limits_R f(x,y)\, dA = \int_a^b \text{Area}(x)\, dx = \int_a^b\int_{g(x)}^{h(x)} f(x,y)\, dy\, dx.\]

Let \(R\) be the type II region lying between \(x=g(y)\) and \(x=h(y)\) for \(a\le y\le b\). Then

\[\iint\limits_R f(x,y)\, dA = \int_a^b\int_{g(y)}^{h(y)} f(x,y)\, dx\, dy.\]

This situation is analogous to the one above for type I regions, but now our slices are at fixed \(y\)-values. The slice at \(y\) begins at \(g(y)\) and ends at \(h(y)\):

The area under this slice is \[\text{Area}(y)=\int_{g(y)}^{h(y)} f(x,y)\, dx.\] As these slices of area range through from \(y=a\) to \(y=b\), we sweep through the entire volume:

Hence \[\iint\limits_R f(x,y)\, dA = \int_a^b \text{Area}(y)\, dy = \int_a^b\int_{g(y)}^{h(y)} f(x,y)\, dx\, dy.\]

Let us first look at this region \(R\) in the \(xy\)-plane:

The intersection points occur when \(x^3=\sqrt{x}\), which we can solve to find \(x=0\) and \(x=1\). Notice that for \(0\le x\le 1\), the curve \(y=\sqrt{x}\) is on top and \(y=x^3\) is on the bottom. Hence \(R\) is the type I region defined by \(0\le x\le 1\) and \(x^3\le y\le \sqrt{x}\).

We can now write the desired double integral as an iterated integral: \[\iint\limits_R x^2 y\, dA = \int_0^1\int_{x^3}^{\sqrt{x}} x^2 y\, dy\, dx\] We evaluate the inner integral treating \(x\) as constant: \[\begin{aligned}\int_{x^3}^{\sqrt{x}} x^2 y\, dy &= \frac{1}{2}x^2y^2 \Big|_{x^3}^{\sqrt{x}} \\ &= \frac{1}{2}(x^3-x^8) \end{aligned}\]

Using this, we have \[\begin{aligned} \iint\limits_R x^2 y\, dA &= \int_0^1\int_{x^3}^{\sqrt{x}} x^2 y\, dy\, dx \\ &= \int_0^1 \frac{1}{2}(x^3-x^8)\, dx \\ &= (x^4/8 – x^9/18) \Big|_0^1 \\ &= (1/8-1/18)-(0-0) = 5/72. \end{aligned}\]

Our inner integral \(\int_{x^3}^{\sqrt{x}} x^2y\, dy\) gives the area of the slice at any \(x\) value. For example, at \(x=0.6\), we have the slice shown below.

Plugging \(x=0.6\) into this integral, we see that the area of this slice is \[\begin{aligned} \int_{0.6^3}^{\sqrt{0.6}} 0.6^2y\, dy &= \int_{0.216}^{\sqrt{0.6}} 0.36y\, dy \\ &= 0.18y^2\Big|_{0.216}^{\sqrt{0.6}} \\ &= 0.18(0.6 – 0.216^2) = 0.0996. \end{aligned}\] Note that this is exactly what we get when we plug \(x=0.6\) into \(\frac{1}{2}(x^3-x^8)\), as we computed in general that \[\int_{x^3}^{\sqrt{x}} x^2y\, dy = \frac{1}{2}(x^3-x^8).\]

As we let \(x\) range from \(0\) to \(1\), these slices sweep out the entire volume of \(5/72\):

All graphics on this page were generated by the Mathematica notebook 15_3DoubleIntegralsOverGeneralRegions.

This notebook generates images and animations like those on this page for any type I region with \(y=g(x)\) and \(y=h(x)\). The notebook also provides many additional options, including the option to attempt to treat the region as type II.

As an exercise, use the notebook to provide a clear graphical answer to Question 3. You can do this simply by setting regionType = 2 and re-executing the code in the Graphics section. Then feel free to change the bounds and the function \(f(x,y)\) itself, keeping in mind the limitations described in the notebook.