I'm homelearning double integrals and currently trying to learn how to calculate the area using double integrals.

I'm trying to solve the following problem:

We have a area bounded by 4 curves: $y=\frac{1}{x}, y^2=x, y=2, x=0$. Calculate it's area.

Could you please help me determine which integrals to calculate? I know how to do it with single variable integrals, but I'm not sure how to define the integrals in multivariable calculus.


  • $\begingroup$ you have $0\le x\le\frac{1}{y}$ and $\sqrt{x}\le y\le 2$ $\endgroup$ – Henry Lee Jun 2 '20 at 16:10
  • $\begingroup$ @HenryLee That's the boundaries, but how do I get the functions to integrate? $\endgroup$ – Daniel Jun 2 '20 at 16:14
  • $\begingroup$ Is you able to draw graph with given curves ? $\endgroup$ – ਮੈਥ Jun 2 '20 at 16:14

The area that you are interested in is the area bounded by the $4$ thick lines from the next picture:

enter image description here

There are points $(x,y)$ in that region with $y$ taking any value from $0$ to $2$. For every such $y$, the values that $x$ can take go from $0$ to:

  • $\sqrt x$ if $x\in[0,1]$;
  • $\frac1x$ if $x\in[1,2]$.

So, compute$$\int_0^1\int_0^{\sqrt x}1\,\mathrm dx\,\mathrm dy+\int_1^2\int_0^{1/x}1\,\mathrm dx\,\mathrm dy.$$

  • $\begingroup$ Thanks, so I use $1$ as function and use my curves as boundaries. $\endgroup$ – Daniel Jun 2 '20 at 16:48
  • $\begingroup$ I have no idea about what it is to “use $1$ as function”. $\endgroup$ – José Carlos Santos Jun 2 '20 at 16:55
  • $\begingroup$ I used it as a number. $\endgroup$ – José Carlos Santos Jun 2 '20 at 17:02

The integrand is 1 in double integral. The area is calculated by summing up all the small areas $\mathrm dS = \mathrm dx \mathrm dy$.

To proceed, we cut the bounded area into two parts, and there are two common plans to do this.

Plan A: Cut the area with $x=1/2$ (image). $$ \int^{1/2}_0 \mathrm dx \int_{\sqrt x}^2 \mathrm dy + \int^{1}_{1/2} \mathrm dx \int_{\sqrt x}^{1/x} \mathrm dy. $$

Plan B: Cut the area with $y=1$ (image). $$ \int^{1}_0 \mathrm dy \int_{0}^{y^2} \mathrm dx + \int^{2}_1 \mathrm dy \int_{0}^{1/y} \mathrm dx. $$

Both give you the answer $\color{Green}{\ln 2 + 1/3}$.


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