# Find the area of the region bounded by the curve $y={{x}^{2}}$ and the line y = 4.

Last updated date: 25th Mar 2023

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Answer

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Hint: The curve $y={{x}^{2}}$ is symmetrical to the y-axis and the line y = 4 restricts the region bounded by a parabola $y={{x}^{2}}$. So, we will first find the area of a region bounded by the y-axis, curve $y={{x}^{2}}$ in the first quadrant and the line y = 4, and then multiply it by 2 to get the required area.

Complete step-by-step answer:

It is given in the question that the region is bounded by the curve $y={{x}^{2}}$ and the line y = 4.

We have to find out the area bounded by the curve $y={{x}^{2}}$ and the line y = 4.

Let us assume that the line AB represents y = 4 in the diagram and curve AOB represent the curve $y={{x}^{2}}$ in the diagram.

Now, from the figure, we can see that we have to find the area of the region AOBA.

From the diagram, we can say that AOB is a parabola which is symmetrical about the y-axis.

So, area of region AODA = area of region BODB.

Also, we have to find the area bounded by the region AOBA.

So, area of the region (AOBA) = 2 x Area(BODB).

Also, area of region (BODB),

$=\int_{0}^{4}{xdy}..........\left( 1 \right)$

We have given that in the question,

$\begin{align}

& {{x}^{2}}=y \\

& \Rightarrow x=\pm \sqrt{y} \\

\end{align}$

As we are finding the area BODB which is lying in the first quadrant. So, we will consider only $x=+\sqrt{y}$ and neglect $x=-\sqrt{y}$ because the first quadrant has positive abscissa and positive ordinate.

So, putting the value of $x=+\sqrt{y}$ in equation (1), we get,

$\begin{align}

& =\int_{0}^{4}{\sqrt{y}}dy \\

& =\int_{0}^{4}{{{\left( y \right)}^{\dfrac{1}{2}}}.}\ dy...........\left( 2 \right) \\

\end{align}$

We know that from basic integration,

$\int_{0}^{a}{{{\left( y \right)}^{n}}.\ dy=\left[ \dfrac{{{y}^{n+1}}}{n+1} \right]_{0}^{a}}$

Applying basic integration on equation (2), we get,

$\begin{align}

& =\left[ \dfrac{{{y}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]_{0}^{4} \\

& =\left[ \dfrac{{{y}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]_{0}^{4} \\

\end{align}$

Now, putting the value of given limits, we get,

$\begin{align}

& =\left[ \dfrac{{{\left( 4 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]-\left[ \dfrac{{{0}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right] \\

& =\left[ \dfrac{{{\left( {{\left( 2 \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]=0 \\

& =\dfrac{{{\left( 2 \right)}^{3}}}{\dfrac{3}{2}} \\

& =\dfrac{8}{\dfrac{3}{2}} \\

& =\dfrac{8\times 2}{3} \\

& =\dfrac{16}{3} \\

\end{align}$

So, the area bounded by the region $BODB=\dfrac{16}{3}$.

Also, the area bounded by the region (AOBA) = 2 x (area of BODB)

Now, putting the value of the area of region BODB, we get,

$\begin{align}

& =2\times \dfrac{16}{3} \\

& =\dfrac{32}{3} \\

\end{align}$

Thus, the area bounded by the region $AOBA=\dfrac{32}{3}$.

Note: The alternative method of this question will be:

Step 1: We will find the area bounded by the region BODB.

Step 2: We will find the area bounded by the region AODA.

Step 3: Adding the area of AODA and area of BODA to get the required area AOBA.

Complete step-by-step answer:

It is given in the question that the region is bounded by the curve $y={{x}^{2}}$ and the line y = 4.

We have to find out the area bounded by the curve $y={{x}^{2}}$ and the line y = 4.

Let us assume that the line AB represents y = 4 in the diagram and curve AOB represent the curve $y={{x}^{2}}$ in the diagram.

Now, from the figure, we can see that we have to find the area of the region AOBA.

From the diagram, we can say that AOB is a parabola which is symmetrical about the y-axis.

So, area of region AODA = area of region BODB.

Also, we have to find the area bounded by the region AOBA.

So, area of the region (AOBA) = 2 x Area(BODB).

Also, area of region (BODB),

$=\int_{0}^{4}{xdy}..........\left( 1 \right)$

We have given that in the question,

$\begin{align}

& {{x}^{2}}=y \\

& \Rightarrow x=\pm \sqrt{y} \\

\end{align}$

As we are finding the area BODB which is lying in the first quadrant. So, we will consider only $x=+\sqrt{y}$ and neglect $x=-\sqrt{y}$ because the first quadrant has positive abscissa and positive ordinate.

So, putting the value of $x=+\sqrt{y}$ in equation (1), we get,

$\begin{align}

& =\int_{0}^{4}{\sqrt{y}}dy \\

& =\int_{0}^{4}{{{\left( y \right)}^{\dfrac{1}{2}}}.}\ dy...........\left( 2 \right) \\

\end{align}$

We know that from basic integration,

$\int_{0}^{a}{{{\left( y \right)}^{n}}.\ dy=\left[ \dfrac{{{y}^{n+1}}}{n+1} \right]_{0}^{a}}$

Applying basic integration on equation (2), we get,

$\begin{align}

& =\left[ \dfrac{{{y}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]_{0}^{4} \\

& =\left[ \dfrac{{{y}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]_{0}^{4} \\

\end{align}$

Now, putting the value of given limits, we get,

$\begin{align}

& =\left[ \dfrac{{{\left( 4 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]-\left[ \dfrac{{{0}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right] \\

& =\left[ \dfrac{{{\left( {{\left( 2 \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]=0 \\

& =\dfrac{{{\left( 2 \right)}^{3}}}{\dfrac{3}{2}} \\

& =\dfrac{8}{\dfrac{3}{2}} \\

& =\dfrac{8\times 2}{3} \\

& =\dfrac{16}{3} \\

\end{align}$

So, the area bounded by the region $BODB=\dfrac{16}{3}$.

Also, the area bounded by the region (AOBA) = 2 x (area of BODB)

Now, putting the value of the area of region BODB, we get,

$\begin{align}

& =2\times \dfrac{16}{3} \\

& =\dfrac{32}{3} \\

\end{align}$

Thus, the area bounded by the region $AOBA=\dfrac{32}{3}$.

Note: The alternative method of this question will be:

Step 1: We will find the area bounded by the region BODB.

Step 2: We will find the area bounded by the region AODA.

Step 3: Adding the area of AODA and area of BODA to get the required area AOBA.

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