Question

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

Answer 1

I got C. Can someone confirm it for me?

Answer 2

How would you solve for y?

Answer 3

Take the square root of both sides.

Answer 4

Anyone?

Answer 5

@jim_thompson5910

Answer 6

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

Answer 7

\[y^2=x \left( 3-x \right)=3x-x^2=-\left( x^2-3x \right)=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\left( \frac{ -3 }{ 2 } \right)^2 \right)\]
\[y^2=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\frac{ 9 }{ 4 } \right)\]
\[y^2=-\left( x-\frac{ 3 }{ 2 } \right)^2+\frac{ 9 }{ 4 }\]
\[y^2+\left( x-\frac{ 3 }{ 2 } \right)^2=\left( \frac{ 3 }{ 2 } \right)^2\]
it is a circle with center (3/2,0) and radius 3/2

area\[=2\int\limits_{0}^{3}\sqrt{x \left( 3-x \right)}dx\]
\[area=2\int\limits_{0}^{3}\sqrt{3x-x^2}dx\]