# express the area in terms of an integral

• Aug 9th 2010, 12:49 PM
dat1611
express the area in terms of an integral
express the area of the region in the x,y plane by the curve x^2/9+y^2/4=1 in terms of an integral

i solved the equation for y to get

y=sqrt(4-2/3x^2)

and looking at the graph i think it would be
3
∫sqrt(4-2/3X^2)
-3

can anyone tell me if this is at all right
• Aug 9th 2010, 01:16 PM
Quote:

Originally Posted by dat1611
express the area of the region in the x,y plane by the curve x^2/9+y^2/4=1 in terms of an integral

i solved the equation for y to get

y=sqrt(4-2/3x^2)

and looking at the graph i think it would be
3
∫sqrt(4-2/3X^2)
-3

can anyone tell me if this is at all right

$\displaystyle \displaystyle\huge\frac{x^2}{9}+\frac{y^2}{4}=1$

$\displaystyle \displaystyle\huge\frac{4}{9}x^2+y^2=4$

$\displaystyle y^2=4-\frac{4}{9}x^2$

$\displaystyle y=\pm\sqrt{4-\frac{4}{9}x^2}=\sqrt{4-\left(\frac{2}{3}x\right)^2}$

Looking at your original function, it's the equation of an ellipse,
hence half of the graph lies below the x-axis.
It's symmetrical, so you should double the integral, since you have only allowed for positive y.

$\displaystyle \displaystyle\huge\frac{x^2}{9}+\frac{y^2}{4}=1$

$\displaystyle x=0,\ y=\pm2$

$\displaystyle y=0,\ x=\pm3$

$\displaystyle Area=2\displaystyle\huge\int_{x=-3}^{x=3}\sqrt{4-\left(\frac{2}{3}x\right)^2}\ dx$