# Calculating area using double integrals

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.

Thanks

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

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

• Thanks, so I use $1$ as function and use my curves as boundaries. – Daniel Jun 2 '20 at 16:48
• I have no idea about what it is to “use $1$ as function”. – José Carlos Santos Jun 2 '20 at 16:55
• I used it as a number. – 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 $$\color{Green}{\ln 2 + 1/3}$$.