double integral bounds question

**Frostking** · March 14th 2009, 11:05 PM

I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

S is the region where x^2 + y^2 is greater or equal to 4
and where x^2 + y^2 is less than or equal to 25
y is greater or equal to 0

I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

I can not see why theta goes from 0 to pi.

Once the bounds are set this is a ridiculously easy integral to take I know!!!!

I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking

**Chris L T521** · March 14th 2009, 11:50 PM

Originally Posted by Frostking

I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

S is the region where x^2 + y^2 is greater or equal to 4
and where x^2 + y^2 is less than or equal to 25
y is greater or equal to 0

I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

I can not see why theta goes from 0 to pi.

Once the bounds are set this is a ridiculously easy integral to take I know!!!!

I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking

Since you are told that y is greater than zero, you're considering all values above the x axis. Since the x-axis is a horizontal line, it has a degree measure of 180 or $\pi$ radians. That is why $0\leqslant\theta\leqslant\pi$ .

**DeMath** · March 15th 2009, 05:06 AM

Originally Posted by Frostking

I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

S is the region where x^2 + y^2 is greater or equal to 4
and where x^2 + y^2 is less than or equal to 25
y is greater or equal to 0

I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

I can not see why theta goes from 0 to pi.

Once the bounds are set this is a ridiculously easy integral to take I know!!!!

I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking

$S = {S_2}\backslash {S_1}$ , where

${S_1} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 4,{\text{ }}y \geqslant 0,{\text{ }} - 2 \leqslant x \leqslant 2} \right\}$ $= \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {4 - {x^2}} ,{\text{ }} - 2 \leqslant x \leqslant 2} \right\}$
and
${S_2} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }} - 5 \leqslant x \leqslant 5} \right\}$ $= \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 5 \leqslant x \leqslant 5} \right\}.$

$\iint\limits_S {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \iint\limits_{{S_2}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} - \iint\limits_{{S_1}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} =$ $\int\limits_{ - 5}^5 {\int\limits_0^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} } - \int\limits_{ - 2}^2 {\int\limits_0^{\sqrt {4 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} } .$

Look a projection of the integration's area on the plane $XY$ on this picture

Your answer must be $3\pi$ .

**DeMath** · March 15th 2009, 11:44 AM

You can provide a projection area of integration in the XY-plane and thus

$S = {S_1} \cup 2{S_2}$ , where

${S_1} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \geqslant 4,{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }} - 2 \leqslant x \leqslant 2,} \right\} =$ $^{}\left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}\sqrt {4 - {x^2}} \leqslant {\text{ }}y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 2 \leqslant x \leqslant 2} \right\}$
and
${S_2} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }}2 \leqslant x \leqslant 5,} \right\} =$ $^{\text{ }}\left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }}2 \leqslant x \leqslant 5} \right\}.$

Then we have

$\iint\limits_S {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \iint\limits_{{S_1}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} + 2\iint\limits_{{S_2}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} =$ $\int\limits_{ - 2}^2 {\int\limits_{\sqrt {4 - {x^2}} }^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} } + 2\int\limits_2^5 {\int\limits_0^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }} = \ldots = 3\pi } } .$

Look this picture

Thread: double integral bounds question

LinkBack

Thread Tools

Display

double integral bounds question

Similar Math Help Forum Discussions

Double integral in polar coordinates- funky bounds!?

How did they get these bounds for this double integral?

Triple integral bounds question

Help with the bounds on this double integral

Finding the bounds for double integrals

Search Tags