# Thread: Area enclosed by two curves

1. ## Area enclosed by two curves

Is my solution correct

QUESTION:
Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
and $\displaystyle x=2+y^2$

this is my solution, i ommited the intermidiete steps...
coverted them to a more familiar functions of y
$\displaystyle y=\frac{3}{x}$
$\displaystyle y = \sqrt{x - 2}$

the curves intercept at the two points, (3,1) and (6,2),

$\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx$
$\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx$
= 2.58722??

2. Originally Posted by bijosn
Is my solution correct

QUESTION:
Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
and $\displaystyle x=2+y^2$

this is my solution, i ommited the intermidiete steps...
coverted them to a more familiar functions of y
$\displaystyle y=\frac{3}{x}$
$\displaystyle y = \sqrt{x - 2}$

the curves intercept at the two points, (3,1) and (6,2),

$\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx$
$\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx$
= 2.58722??
$\displaystyle y=\displaystyle \frac{x}{3}$ and $\displaystyle y=\displaystyle \sqrt{x-2}$

$\displaystyle A=\displaystyle \int_3^6(\sqrt{x-2}-\frac{x}{3})dx$

= $\displaystyle \displaystyle \frac{(x-2)^{3/2}}{3/2}-\frac{x^2}{6})\big|_3^{6}$

= $\displaystyle \displaystyle \frac{2}{3}(4^{3/2}-1)-\frac{1}{6}\times 27$

= $\displaystyle \displaystyle \frac{2}{3}\times 7-\frac{9}{2}$

= $\displaystyle \displaystyle \frac{28-27}{6}$

= $\displaystyle \displaystyle \frac{1}{6}$

3. Originally Posted by bijosn
Is my solution correct

QUESTION:
Calculate the area of the region enclosed by the curves$\displaystyle x = 3y$
and $\displaystyle x=2+y^2$

this is my solution, i ommited the intermidiete steps...
coverted them to a more familiar functions of y
$\displaystyle y=\frac{3}{x}$
$\displaystyle y = \sqrt{x - 2}$

the curves intercept at the two points, (3,1) and (6,2),

$\displaystyle \int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx$
$\displaystyle \int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx$
= 2.58722??
no.

$\displaystyle 3y = 2 + y^2$

$\displaystyle y = 2$ , $\displaystyle y = 1$

$\displaystyle \displaystyle A = \int_1^2 3y - (2+y^2) \, dy$

try again.

4. Originally Posted by alexmahone
$\displaystyle y=\displaystyle \frac{x}{3}$ and $\displaystyle y=\displaystyle \sqrt{x-2}$

$\displaystyle A=\displaystyle \int_3^6(\sqrt{x-2}-\frac{x}{3})dx$

= $\displaystyle \displaystyle \frac{(x-2)^{3/2}}{3/2}-\frac{x^2}{6})\big|_3^{6}$

= $\displaystyle \displaystyle \frac{2}{3}(4^{3/2}-1)-\frac{1}{6}\times 27$

= $\displaystyle \displaystyle \frac{2}{3}\times 7-\frac{9}{2}$

= $\displaystyle \displaystyle \frac{28-27}{6}$

= $\displaystyle \displaystyle \frac{1}{6}$
thank you, I made a silly mistake with the first equation, putting it as $\displaystyle y=\frac{3}{x}$ instead of $\displaystyle y =\frac{x}{3}$

5. Originally Posted by skeeter
no.

$\displaystyle 3y = 2 + y^2$

$\displaystyle y = 2$ , $\displaystyle y = 1$

$\displaystyle \displaystyle A = \int_1^2 3y - (2+y^2) \, dy$

try again.
thank you

$\displaystyle \displaystyle A = \int_1^2 3y - 2 - y^2 \, dy$
$\displaystyle \displaystyle = ( \frac{3y^2}{2} - 2y - \frac{y^3}{3} ) \big|_1^{2}$
$\displaystyle = \frac{1}{6}$