# Find the area of the region enclosed

• August 5th 2009, 06:43 AM
dat1611
Find the area of the region enclosed
• August 5th 2009, 06:53 AM
O113
$
dA = f(x) - g(x) <=> A = \int {f(x) - g(x)}\$
where f(x) is the top-function

So, figure out what function is above the other and get the anti derivative, then put in your limits
• August 5th 2009, 07:14 AM
dat1611
so the anti derivative is -3cos(x)-2sin(x)
what do i do after i plug in the limits
• August 5th 2009, 07:44 AM
skeeter
first find the intersection of the two functions in the given interval ...

$3\sin{x} = 2\cos{x}$

$\tan{x} = \frac{2}{3}$

$x = \arctan\left(\frac{2}{3}\right)$

finding the area will consist of two integrals because the "top" function changes ...

let $c = \arctan\left(\frac{2}{3}\right)$

$A = \int_0^c 2\cos{x} - 3\sin{x} \, dx + \int_c^{\frac{7\pi}{10}} 3\sin{x} - 2\cos{x} \, dx
$

integrate and use the Fundamental Theorem of Calculus to find the area value.

note that this area problem is probably better suited for completion with a calculator.
• August 5th 2009, 09:25 AM
DeMath
So you have ${y_1} = 3\sin x,{\text{ }}{y_2} = 2\cos x,{\text{ }}0 \leqslant x \leqslant \frac{{7\pi }}{{10}},{\text{ }}S = ?$

${y_1} = {y_2} \Leftrightarrow 3\sin x = 2\cos x \Leftrightarrow \tan x = \frac{2}{3} \Leftrightarrow x = \arctan \frac{2}{3} + \pi k{\text{ }}\left( {k \in \mathbb{Z}} \right).$

$S = {S_1} + {S_2}.$

${S_1} = \int\limits_0^{\arctan \frac{2}{3}} {\left( {2\cos x - 3\sin x} \right)dx} = \left. {\left( {2\sin x + 3\cos x} \right)} \right|_0^{\arctan \frac{2}{3}} =$

$= 2\sin \left( {\arctan \frac{2}{3}} \right) + 3\cos \left( {\arctan \frac{2}{3}} \right) - 3 =$

$= 2 \cdot \frac{{{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}}}{{\sqrt {1 + {{\left( {{2 \mathord{\left/
{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} + 3 \cdot \frac{1}{{\sqrt {1 + {{\left( {{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} - 3 =$

$= \frac{4}{{\sqrt {13} }} + \frac{9}{{\sqrt {13} }} - 3 = \frac{{13}}{{\sqrt {13} }} - 3 = \sqrt {13} - 3.$

${S_1} = \int\limits_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} {\left( {3\sin x - 2\cos x} \right)dx} = - \left. {\left( {3\cos x + 2\sin x} \right)} \right|_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} =$

$= - \left( {3\cos \frac{{7\pi }}{{10}} + 2\sin \frac{{7\pi }}{{10}} - \sqrt {13} } \right) = \sqrt {13} - 3\cos \left( {\pi - \frac{{3\pi }}{{10}}} \right) - 2\sin \left( {\pi - \frac{{3\pi }}{{10}}} \right) =$

$= \sqrt {13} + 3\cos \frac{{3\pi }}{{10}} - 2\sin \frac{{3\pi }}{{10}} = \sqrt {13} + 3 \cdot \frac{{\sqrt {5 - \sqrt 5 } }}{{2\sqrt 2 }} - 2 \cdot \frac{{\sqrt 5 + 1}}{4} =$

$= \sqrt {13} + \frac{{3\sqrt 2 \sqrt {5 - \sqrt 5 } }}{4} - \frac{{\sqrt 5 + 1}}{2} = \sqrt {13} + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{1}{2}\left( {\sqrt 5 + 1} \right).$

Finally

$S = \sqrt {13} - 3 + \sqrt {13} + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{1}{2}\left( {\sqrt 5 + 1} \right) =$

$= \frac{1}{2}\left( {4\sqrt {13} - \sqrt 5 } \right) + \frac{3}{4}\sqrt {10 - 2\sqrt 5 } - \frac{7}{2} \approx {\text{4}}{\text{.35642}}$

See this picture