Calculating the area with simple and double integrals

Question

I have an exam on calculus next week and I'm confused with the usage of simple and double integrals. As far as I know, if you want to calculate the area under the curve of a function $f(x)$ between $a$ and $b$, you have to calculate an integral between those two points: $$\int_{a}^{b}f(x)dx$$

If you want to calculate the integral between those points having as upper and lower boundaries two functions, namely $f(x)$ and $g(x)$, you have to calculate the integral of the subtraction of those functions: $$(1)\int_{a}^{b}(f(x)-g(x))dx$$ where $f(x)$ is the upper function and $g(x)$ the lower one.

However, I'm confused with double integrals because I've been told that by calculating a double integral of a function, you're calculating the volume under the curve of that function. It makes sense to me, but I've seen some examples in which my teacher calculates the area between one or more functions by using a double integral.

Another thing that I don't understand is if I'm calculating the double integral of a function inside a given domain what am I really calculating? For example, if I'm calculating $\iint_{D}\sin (y^{3})\text{dx dy}$ in the following domain:

where $y=\sqrt{x}$ and $x\in [0,1]$. What will that integral give me?

In relation to that, if I just simply calculate $\iint_{D}dxdy$, how will it make a difference? Will I calculate the whole area of D?

And my last question has to do with the formula (1). As I previously said, sometimes my teacher uses double integrals to calculate the area between two functions. Then, don't I need to use formula (1)? Are they perhaps equivalent? For example, given the domain:

$$D=\left \{ (x,y)\in \mathbb{R}^{2}:(y-x^{2}-1)(y-2x^{3})<0,x\in[0,2] \right \}$$

In order to calculate its area, she would split the problem into double integrals, one for $y-x^{2}-1=0$ and another one for $y-2x^{3}=0$, and find the points where they intersect:

Like this:

$$\iint_{D}dxdy = \iint_{D_{1}}dxdy + \iint_{D_{2}}dxdy$$

Can't I use my formula (1) in this case?

Thank you so much in advance.

Answer 1

By a double integral in the form

$$\iint_{D}f(x,y)dxdy$$

we are evaluating the (signed) volume between the function $z=f(x,y)$ and the x-y plane.

Then by the following double integral

$$\iint_{D}1\cdot dxdy$$

we are evaluating the volume of a cilinder with base the domain $D$ and height $1$ which correspond (numerically) to the area of the domain $D$.

For the last question, yes the integral is additive thus we can divide it as

$$\iint_{D}dxdy = \iint_{D_{1}}dxdy + \iint_{D_{2}}dxdy$$

and we can also use (1) splitting the integral accordingly.

Answer 2

The area of $D$ can be calculated by both $\iint_D dx dy$ and by $\int_0^1 (1-\sqrt{x}) dx$.

The way I like to think about it is in terms of "dimensional analysis". In the double integral, you're integrating $1$ across the domain of $D$, so it gives you the area of $D$. In essence, that $dx dy$ bit tells you that you are integrating two dimensions, kinda like multiplying two dimensions. So it gives you an area. Similarly, you have the height $(1-\sqrt{x})$ for a given infinitesimal length $dx$. When multiplied and added together, you get an area, because a length times a width is an area.

Another way I like to think about it is by thinking about putting little spaghetti strands for $dx$ or $dy$. If you do it the "double-integral" way, you're standing up spaghetti strands of size $1$ on each tiny $dxdy$ square. The resulting volume is the area of $D$ times 1 "spaghetti height". For the single integral, you lay down spaghetti of length $(1- \sqrt{x_0})$ at $x = x_0$, doing that in each part of $D$. The resulting area is the area of $D$.

So, they are two different integrals, with two different interpretations. The way you want to represent it largely depends on the problem you're trying to solve, the techniques you know, and how you want to communicate your results.

EDIT: Using the spaghetti method, imagine spaghetti of height $\sin(y^3)$ for each square, so it would be $y$-dependent. You can sorta see the wavy-ness of the sin wave differ with $y$. You do need to calculate $\iint_D \sin(y^3) dy dx$, which is different from $\iint_D dy dx =$ area of $D$. You can use equations as your endpoints; I usually set it up like $$\int_0^1\int_{1-\sqrt{x}}^1 \sin(y^3) dy dx$$ and then I solve it inside-out.