Double integrals over general areas

Connecting area and volume views

Or...connecting "area between curves" and "volume of a solid" interpretations of double-integrals.

Consider the green area in the $xy$ plane

According to the "area between curves" view, the area in green is given by: $$\begineq A &=& \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} dA\\ &=&\int_{x=0}^{1}\left(\int_{y=1-x}^{\sqrt{1-x^2}} \,dy\right)\,dx\\ &=&\int_{x=0}^{1}\left(\left. y\right|_{y=1-x}^{\sqrt{1-x^2}}\right)\,dx\\ &=&\int_{x=0}^{1}\left(\sqrt{1-x^2}-(1-x)\right) \,dx\\ &=&\frac{\pi}{4} -\frac 12 \endeq $$

But we could instead think about that double integral as the volume above a non-rectangular area in the $xy$ plane, going up to the surface $f(x,y)=1$, like this... $$\begineq V &=& \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} dA\\ &=& \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} 1\,dA\\ &=& \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} f(x,y)\,dA\\ \endeq$$

Then the volume corresponds to this solid:

  • The base in the $xy$ plane is the green area $A$,
  • it extends straight up from $A$,
  • to a height of $1$...
  • that is, it is bounded above by the surface $f(x,y)=1$.

The value of the volume integral: $$ V = \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} 1 \,dy\,dz$$ is the same as the value of the area between curves that we first considered.

Double integrals above general areas

$$\iint_A f(x,y)\,dA=?$$ Let's generalize our interpretation of this integral as the volume under $f(x,y)=1$ to *any* function $f(x,y)$: In general this double integral is...

  • The volume of a solid,
  • bounded, on the bottom, by a non-rectangular area, $A$, in the $xy$ plane.
  • going straight up from that area,
  • bounded above by a surface $f(x,y)$

This volume integral is represented in the picture below: $$V = \int_{x=0}^{1}\int_{y=1-x}^{\sqrt{1-x^2}} x\,dA$$
It's bounded by $f(x,y)=x$ above.

We shall find that now, the order of integration *does* affect how we write down the limits of a double integral.

Order matters.

$\iint_A f(x,y)\,dA$

  1. Integrate first with respect to y
  2. This gives us the area, $A(x)$, of a slice through the solid above the blue arrow, which depends on $x$, $$A(x)=\int_{y=1-x}^{\sqrt{1-x^2}} f(x,y)\,dy.$$
  3. Integrate second with respect to x

Written... $$\int_{x=0}^1A(x)\,dx=\int_{x=0}^1\int_{y=1-x}^{\sqrt{1-x^2}} f(x,y)\,dy\,dx$$

Order matters..

$\iint_A f(x,y)\,dA$

  1. Integrate first with respect to x,
  2. This gives us the area, $A(y)$, of a slice through the solid above the red arrow, which depends on $y$, $$A(y)=\int_{x=1-y}^{\sqrt{1-y^2}} f(x,y)\,dx$$
  3. Integrate second with respect to y

Written... $$\int_{y=0}^1A(y)\,dy=\int_{y=0}^1\int_{x=1-y}^{\sqrt{1-y^2}} f(x,y)\,dx\,dy.$$

Example

$\iint_A (4x+2)\,dA$

Regions

You should be able to sketch the region of integration in $x$ and $y$ given a double integral. For example: $$\int_0^1\int_y^\sqrt{y} x^2y^2\,dx\,dy$$

...means $$\int_{y=0}^1\left(\int_{x=y}^\sqrt{y} x^2y^2\,dx\right)\,dy$$

$\Rightarrow$

Changing the order of integration

$$\int_{x=-1}^0\int_{y=0}^{x+1}e^{x+y}dy\,dx+\int_{x=0}^1\int_{y=0}^{1-x} e^{x+y}dy\,dx$$

$\Rightarrow$

We could get the same result in one integral, integrating first over $x$ then $y$ like this...

The line $y=x+1$ can be re-arranged to $x=y-1$.
The line $y=1-x$ can be re-arranged to $x=1-y$: $$\int_{y=0}^1\int_{x=y-1}^{1-y} e^{x+y}dx\,dy$$

To do

  • Double Integrals Practice: #5-7
  • Limits on Double Integrals
  • Double Integrals: Problems