# Thread: double integral bounds question

1. ## double integral bounds question

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

2. 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 $0\leqslant\theta\leqslant\pi$.

3. 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 $3\pi$.

4. 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