Question

The upper half of an ellipse centered at the origin with axes length 6 and 4 is given by... find the area of the entire ellipse.

Answer 1

Are you in calculus?

Answer 2

If so find the area under the curve and multiply by 2.

\[2/3\int\limits_{-3}^{3}\sqrt{9+x ^{2}}dx\]

Answer 3

yeah. arent the bounds from 4 to 6?

Answer 4

I'm assuming that by "axis" the entire horiz. axis (not just half of it) is meant. In that case the bounds for x would be -2 to 2 and for y -3 to 3.

Draw a picture.

Ask if you have further questions about this problem.

Answer 5

There's some confusion here as to whether the ellipse is a vertical or horizontal one.
You may have to examine the function f(x) carefully to determine which is which.

Answer 6

it looks like it's a vertical ellipse ?

Answer 7

\[y=\frac{ 2 }{3 }\sqrt{9-x^2},\frac{ 9 }{ 4 }y^2=9-x^2,x^2+\frac{ 9 }{ 4 }y^2=9\]
divide by 9\[\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 4 }=1\]
major axis is along x-axis
\[reqd.~ area=\int\limits_{-3}^{3}\frac{ 2 }{ 3 }\sqrt{9-x^2}dx\]
when we put x=-x function remains the same.
so it is an even function.
\[reqd. ~area=2 \times \frac{ 2 }{ 3 }\int\limits_{0}^{3}\sqrt{9-x^2}dx\]
\[put ~x=3 \sin \theta,dx=3\cos \theta ~d \theta\]
when x=0
\[\sin \theta=0,\theta=0\]
when x=3
\[\sin \theta=\frac{ 3 }{ 3 }=1,\theta=\frac{ \pi }{ 2 }\]
\[reqd.~area=\frac{ 4 }{ 3 }\int\limits_{0}^{\frac{ \pi }{ 2 }} 3\sqrt{1-\sin ^2\theta}~3~\cos \theta ~d \theta \]
\[=6\int\limits_{0}^{\frac{ \pi }{ 2 }}2~\cos ^2\theta~d \theta =6\int\limits_{0}^{\frac{ \pi }{ 2 }}\left( 1+\cos 2\theta \right)d \theta\]
=?