Question

# Find the area of the ellipse: $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$

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Hint:To find the area of the ellipse, we will convert the given equation in terms of y and then integrate the region under the graph.
Here, the first step is to convert the given equation in terms of $y$.
Therefore the above equation can be written as,
$\frac{{{y^2}}}{9} = 1 - \frac{{{x^2}}}{4}$
If we solve it further, we get,
${y^2} = 9\left( {\frac{{4 - {x^2}}}{4}} \right)$
$y = \frac{3}{2}\left( {\sqrt {4 - {x^2}} } \right)$
Now the next step is to use the formula of area of ellipse which is
Area of ellipse $4$(area of region$OAB$)
=$4\int\limits_0^2 {ydx}$
Note: Make sure you take the right limits.
We are going to equate the value of y.
=$4\int\limits_0^2 {\frac{3}{2}\left( {\sqrt {4 - {x^2}} } \right)dx}$
Now we take the constant out of the integration. And we convert $\sqrt {4 - {x^2}}$to a general form of $\sqrt {{a^2} - {x^2}}$.Therefore we get,
=$4\left( {\frac{3}{2}} \right)\int\limits_0^2 {\left( {\sqrt {{2^2} - {x^2}} } \right)dx}$
We know that the integration of $\sqrt {{a^2} - {x^2}}$ is equal to $\frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + c$
Therefore,
=$6\left[ {\frac{1}{2} \times \sqrt {4 - {x^2}} + \frac{1}{2}\left( 4 \right){{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]$
=$\frac{6}{2}\left[ {0 + 4{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right)} \right]$
=$3\left[ {4{{\sin }^{ - 1}}\left( 1 \right)} \right]$
=$12\left( {\frac{\pi }{2}} \right)$
Answer =$6\pi$