# Thread: Application of integral calculus - 2

1. ## Application of integral calculus - 2

1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

For me these problems have been the most difficult some of them, I began to deal with, but my answers were not correct.
Tell me please, what program would be easier to do the drawings for these tasks.

2. Originally Posted by Dext91

1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

For me these problems have been the most difficult some of them, I began to deal with, but my answers were not correct.
Tell me please, what program would be easier to do the drawings for these tasks.
For number 1

So $\displaystyle r = 7\left( {1 - \sin \phi } \right),{\text{ }} - \frac{\pi }{6} \leqslant \phi \leqslant \frac{\pi }{6},{\text{ }}S = ?$

Solution:

$\displaystyle S = \frac{1} {2}\int\limits_\alpha ^\beta {{r^2}d\phi } = \frac{1} {2}\int\limits_{{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 6}} \right. \kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/ {\vphantom {\pi 6}} \right. \kern-\nulldelimiterspace} 6}} {{{\left( {7\left( {1 - \sin \phi } \right)} \right)}^2}d\phi } =$

$\displaystyle = \frac{{49}} {2}\int\limits_{{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 6}} \right. \kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/ {\vphantom {\pi 6}} \right. \kern-\nulldelimiterspace} 6}} {\left( {1 - 2\sin \phi + {{\sin }^2}\varphi } \right)d\phi } =$

$\displaystyle = \left. {\frac{{49}}{2}\left( {\frac{{3\phi }}{2} + 2\cos \phi } \right)} \right|_{{{ - \pi } \mathord{\left/{\vphantom {{ - \pi } 6}} \right.\kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}}$ $\displaystyle - \frac{{49}} {4}\int\limits_{{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 6}} \right. \kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/ {\vphantom {\pi 6}} \right. \kern-\nulldelimiterspace} 6}} {\cos 2\phi d\phi } =$

$\displaystyle = \left. {\frac{{49}}{2}\left( {\frac{{3\phi }}{2} + 2\cos \phi - \frac{1}{4}\sin 2\phi } \right)} \right|_{{{ - \pi } \mathord{\left/{\vphantom {{ - \pi } 6}} \right.\kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} = \frac{{49}}{2}\left( {2\pi - \sqrt 3 } \right).$

Look this picture

3. ## Answer to second question

See attachment

4. Originally Posted by Dext91

1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

For me these problems have been the most difficult some of them, I began to deal with, but my answers were not correct.
Tell me please, what program would be easier to do the drawings for these tasks.
So $\displaystyle y = 2 + \ln x,{\text{ }}\sqrt 3 \leqslant x \leqslant \sqrt 8 ,{\text{ }}L = ?$
$\displaystyle L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}} {{dx}}} \right)}^2}} dx} =$$\displaystyle \int\limits_{\sqrt 3 }^{\sqrt 8 } {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}\left( {2 + \ln x} \right)} \right)}^2}} dx} = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\sqrt {1 + \frac{1}{{{x^2}}}} dx} = \displaystyle = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{1}{x}\sqrt {{x^2} + 1} dx} = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{{x^2} + 1}}{{x\sqrt {{x^2} + 1} }}dx} = \displaystyle \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x} {{\sqrt {{x^2} + 1} }}dx} + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{1} {{x\sqrt {{x^2} + 1} }}dx} = \displaystyle = \int\limits_{\sqrt 3 }^{\sqrt 8 } {d\left( {\sqrt {{x^2} + 1} } \right) + } \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}{{{x^2}\sqrt {{x^2} + 1} }}dx} = \displaystyle \left. {\sqrt {{x^2} + 1} } \right|_{\sqrt 3 }^{\sqrt 8 } + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}{{\left( {{x^2} + 1 - 1} \right)\sqrt {{x^2} + 1} }}dx} = \displaystyle = 1 + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{{x \mathord{\left/{\vphantom {x {\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}} \right.\kern-\nulldelimiterspace} {\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}}} {{1 - {{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}}}dx} = 1 - \displaystyle \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{d\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}}{{1 - {{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}}}}$$\displaystyle =1 - \left. {\frac{1}{2}\ln \left| {\frac{{1 + \frac{1}{{\sqrt {{x^2} + 1} }}}}{{1 - \frac{1}{{\sqrt {{x^2} + 1} }}}}} \right|} \right|_{\sqrt 3 }^{\sqrt 8 }=$
$\displaystyle = 1 - \left. {\frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} + 1}}{{\sqrt {{x^2} + 1} - 1}}} \right|} \right|_{\sqrt 3 }^{\sqrt 8 }$ $\displaystyle = 1 - \frac{1}{2}\left( {\ln 2 - \ln 3} \right) = 1 + \frac{1}{2}\ln \frac{3}{2}.$