# Area enclosed by two curves

• Feb 6th 2011, 10:46 AM
bijosn
Area enclosed by two curves
Is my solution correct

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

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

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

$\int_3^6\left[ \sqrt{x - 2} - \frac{3}{x} \right] dx$
$\int_3^6 \left[ \left( x - 2 \right)^\frac{1}{2} - 3x^{-1} \right] dx$
= 2.58722??
• Feb 6th 2011, 01:11 PM
alexmahone
Quote:

Originally Posted by bijosn
Is my solution correct

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

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

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

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

$y=\displaystyle \frac{x}{3}$ and $y=\displaystyle \sqrt{x-2}$

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

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

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

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

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

= $\displaystyle \frac{1}{6}$
• Feb 6th 2011, 01:15 PM
skeeter
Quote:

Originally Posted by bijosn
Is my solution correct

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

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

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

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

no.

$3y = 2 + y^2$

$y = 2$ , $y = 1$

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

try again.
• Feb 6th 2011, 08:55 PM
bijosn
Quote:

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

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

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

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

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

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

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

thank you, I made a silly mistake with the first equation, putting it as $y=\frac{3}{x}$ instead of $y =\frac{x}{3}$
• Feb 6th 2011, 09:20 PM
bijosn
Quote:

Originally Posted by skeeter
no.

$3y = 2 + y^2$

$y = 2$ , $y = 1$

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

try again.

thank you

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