## anonymous one year ago The area enclosed by the curve y^2 = x(3 − x) is given by https://gyazo.com/d848359e382f1fcf7369e66b148fc534

1. anonymous

I got C. Can someone confirm it for me?

2. anonymous

How would you solve for y?

3. anonymous

Take the square root of both sides.

4. anonymous

Anyone?

5. anonymous

@jim_thompson5910

6. jim_thompson5910

I agree, you solve for y to get $\Large y = \sqrt{x(3-x)} = \sqrt{3x-x^2}$ you double the area because the integral alone is just figuring out the top portion

7. anonymous

$y^2=x \left( 3-x \right)=3x-x^2=-\left( x^2-3x \right)=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\left( \frac{ -3 }{ 2 } \right)^2 \right)$ $y^2=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\frac{ 9 }{ 4 } \right)$ $y^2=-\left( x-\frac{ 3 }{ 2 } \right)^2+\frac{ 9 }{ 4 }$ $y^2+\left( x-\frac{ 3 }{ 2 } \right)^2=\left( \frac{ 3 }{ 2 } \right)^2$ it is a circle with center (3/2,0) and radius 3/2 |dw:1444436581363:dw| area$=2\int\limits_{0}^{3}\sqrt{x \left( 3-x \right)}dx$ $area=2\int\limits_{0}^{3}\sqrt{3x-x^2}dx$