Question

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

Answer 1

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 =\[6\pi \]