Finding the area enclosed by three curves I need to find the area that is enclosed between three curves. I was trying to look for the smartest solution visually, but haven't gotten anywhere.
Curves:
$$y_1= 2x^2$$
$$y_2 = x^2$$
$$y_3 = \frac{1}{x}$$
Please help
 A: One way to calculate the mentioned area is using double integrals. Strictly speaking, Change of variables theorem works perfect here. The general idea in this theorem is to transform the area to a well-known one, which here seems to be a rectangle, and then apply iteration method. So let the dedicated area be $D$. Then using change of variables theorem for double integrals we have:
$$\text{area of $D$} = \iint\limits_D dx dy = \iint\limits_{D'} \left\lvert \dfrac{\partial(x,y)}{\partial(u,v)}\right\rvert dudv.$$
where $u = \dfrac{y}{x^2}$ and $v = xy$. Note that with new variables, $D$ is transformed to $D'$ which is a rectangle in $\text{$uv$-plane}$ bounded by lines $u=1, u=2$ and $v=1, v=0$. Also note that group of curves $xy = k$ for $0 < k \leq 1$ cover the area of $D$ and each point in $D$ belongs to exactly one of these curves (i.e. the transformation is both surjective and bijective). Having said these and the inverse function theorem which states that $$\left\lvert \dfrac{\partial(x,y)}{\partial(u,v)}\right\rvert = \dfrac{1}{\left\lvert \dfrac{\partial(u,v)}{\partial(x,y)}\right\rvert},$$
we have:
$$\left\lvert \dfrac{\partial(u,v)}{\partial(x,y)}\right\rvert = \dfrac{3y}{x^2} = 3u \Longrightarrow \left\lvert \dfrac{\partial(x,y)}{\partial(u,v)}\right\rvert = \dfrac{1}{3u}.$$
Therfore:
$$\iint\limits_{D'} \left\lvert \dfrac{\partial(x,y)}{\partial(u,v)}\right\rvert dudv = \iint\limits_{D'} \dfrac{1}{3u}dudv = \frac{1}{3} \int_0^1 dv \int_1^2 \dfrac{du}{u} = \frac{1}{3} \times 1 \times (\ln2 - \ln 1) = \frac{1}{3} \ln 2.$$
Further reading at the section 14.4 of Calculus, a complete course by R. A. Adams.
A: 
\begin{align}
\int_0^{1/\sqrt[3]2} f_1(x)&-f_2(x)\, dx
+
\int_{1/\sqrt[3]2}^1 f_3(x)-f_2(x)\, dx
\\
&=
\int_0^{1/\sqrt[3]2} x^2\, dx
+
\int_{1/\sqrt[3]2}^1 \frac1x-x^2\, dx
\\
&=
\tfrac13x^3\Big|_0^{1/\sqrt[3]2}
+
(\ln(x)-\tfrac13x^3)
\Big|_{1/\sqrt[3]2}^1
\\
&=\tfrac13\ln2
\approx 0.2310 
,
\end{align}
A: I'm not sure how to solve this visually, but we can solve it with integrals.
First, we need to find where the different equations intersect each other.

*

*$y_1=y_2\rightarrow2x^2=x^2\rightarrow x^2=0\rightarrow x=0$

*$y_1=y_3\rightarrow2x^2=\frac{1}{x}\rightarrow x^3=\frac{1}{2}\rightarrow x=\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}$

*$y_2=y_3\rightarrow x^2=\frac{1}{x}\rightarrow x^3=1\rightarrow x=1$
The area enclosed by the three curves is $\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}y_1-y_2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1y_3-y_2dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}(2x^2-x^2)dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1(\frac{1}{x}-x^2)dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}x^2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1\frac{1}{x}dx-\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1x^2dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}x^2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1\frac{1}{x}dx-\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1x^2dx=\frac{1}{6}+\ln(\frac{1}{\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}})-\frac{1}{3}+\frac{1}{6}=\frac{1}{3}\ln(2)$
A: After visual observation, the integral is equivalent to
$$\int_0^1\min\left(x^2,\dfrac1x-x^2\right)\,dx$$ which you decompose as
$$\int_0^a x^2\,dx+\int_a^1\left(\dfrac1x-x^2\right)\,dx=\frac{a^3}3-\log a-\frac{1-a^3}3.$$ $a$ is the abscissa of the meeting point of the two functions.
