Answer

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**Hint:**For the given question we are given a parametric equation and asked to find the area enclosed by the graph. By converting given parametric equations to x and y forms we can see that they will be converted into an eclipse graph as we know that the area of any eclipse graph is 4 times of the symmetrical part. Therefore we can find the area of the graph.

**Complete step-by-step answer:**

For the given problem we are given to find the area enclosed by the curve \[x=3\cos t\] and \[y=2\sin t\].

Let us consider given 2 equations as equation (1) and equation (2).

\[x=3\cos t....................\left( 1 \right)\]

\[y=2\sin t...................\left( 2 \right)\]

As we know the identity \[{{\cos }^{2}}t+{{\sin }^{2}}t=1\]. So let us consider the identity as (I1).

\[{{\cos }^{2}}t+{{\sin }^{2}}t=1...........\left( I1 \right)\]

Let us find the value of \[\cos t\] and \[\sin t\] from equation (2) and equation (1) respectively.

From equation (1) and equation (2), we get

\[\dfrac{x}{3}=\cos t\] and \[\dfrac{y}{2}=\sin t\].

Let us substitute the above values in identity (I1), we get

\[\Rightarrow {{\left( \dfrac{x}{3} \right)}^{2}}+{{\left( \dfrac{y}{2} \right)}^{2}}=1\]

Let us consider the above equation as equation (3).

\[\Rightarrow {{\left( \dfrac{x}{3} \right)}^{2}}+{{\left( \dfrac{y}{2} \right)}^{2}}=1.......................\left( 3 \right)\]

Let us plot the graph for the above equation

As we know the equation of eclipse is in the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. So, as we know that the graph is symmetrically partitioned by 4 parts. Therefore, by finding the one part we can find the total area of the graph.

\[\Rightarrow \text{Area=4}\text{.Area of OAB}\]

\[=4\int\limits_{0}^{3}{ydx}\]

Let us consider the above equation as equation (4).

\[A=4\int\limits_{0}^{3}{ydx}...................\left( 4 \right)\]

Therefore, from the equation (1) we get

\[\Rightarrow y=\left( \dfrac{2}{3}\sqrt{\left( 9-{{x}^{2}} \right)} \right)\]

Let us substitute the above value in equation (4) we get,

\[\Rightarrow A=4\int\limits_{0}^{3}{\left( \dfrac{2}{3}\sqrt{\left( 9-{{x}^{2}} \right)} \right)dx}\]

\[\Rightarrow A=\dfrac{8}{3}\int\limits_{0}^{3}{\left( \sqrt{\left( {{3}^{2}}-{{x}^{2}} \right)} \right)dx}\]

Let us consider above equation as equation (5).

\[\Rightarrow A=\dfrac{8}{3}\int\limits_{0}^{3}{\left( \sqrt{\left( {{3}^{2}}-{{x}^{2}} \right)} \right)dx}.........\left( 5 \right)\]

By the formula \[\int{{{a}^{2}}-{{x}^{2}}dx=\left| \dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a} \right|}\]

By applying the formula to the equation (5), we get

\[\Rightarrow A=\dfrac{8}{3}\left| \dfrac{x}{2}\sqrt{{{3}^{2}}-{{x}^{2}}}+\dfrac{9}{2}{{\sin }^{-1}}\dfrac{x}{3} \right|_{0}^{3}\]

\[\Rightarrow A=\dfrac{8}{3}\left| \dfrac{9}{2}{{\sin }^{-1}}\left( \sin \dfrac{\pi }{2} \right) \right|\]

\[\Rightarrow A=6\pi squnits\]

Therefore, the area of the given graph is \[6\pi squints\].

**Note:**While doing this problem we have to remember a point that we have to convert the given parametric equation to the geometric equation to get the graph. We should be aware of all integration formulas to solve this question

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