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# Find the area of the region bounded by the curve $y^2=4x,x^2=4y$.

$\begin{array}{1 1} \frac{16}{2}\;sq.units. \\ \frac{16}{3}\;sq.units. \\ \frac{15}{2}\;sq.units. \\ \frac{15}{3}\;sq.units\end{array}$

Toolbox:
• Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bonded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
• $A=\int_a^b f(x)dx$ or $\int_a^b ydx.$
Step 1:
Given $y^2=4x$ and $x^2=4y$.
Clearly $y^2=4x$ is a parabola.
Focus of the parabola is $4a=4\Rightarrow a=1$
Hence focus=(1,0)
The curve is open rightward.
The curve $x^2=4y$ is a parabola.
Here $4a=4\Rightarrow a=1$
Hence the focus is (0,1)
The curve is open upward.
Step 2:
Let us next find the point of intersection of the curves.
$x^2=4y\Rightarrow y=\large\frac{x^2}{4}$ or $y^2=\large\frac{x^4}{16}$
Now equating this in $y^2=4x$
$\large\frac{x^4}{16}$$=4x\Rightarrow x^3=64. x=4 The point of intersection is (4,0) The required area is the shaded portion in the fig. Step 3: The required area A=\int_0^4(y_2-y_1)dx. Where y_1=\large\frac{x^2}{4} and y_2=\sqrt {4x}. A=2\int_0^4\sqrt x dx-\large\frac{1}{4}\int_0^4$$x^2dx.$
On integrating we get,
$2\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^4$$-\large\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^4=\large\frac{4}{3}\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_0^4-\frac{1}{12}\begin{bmatrix}x^3\end{bmatrix}_0^4 Step 4: On applying limits we get, A=\large\frac{4}{3}[$$4^{\Large\frac{3}{2}}$$-0]-\large\frac{1}{12}[$$4^3-0]$
$\;\;=\large\frac{4\times 8}{3}-\frac{64}{12}$
$\;\;=\large\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$
Hence the required area=$\large\frac{16}{3}$sq.units.
edited Apr 28, 2013