Section 16.3 Double Integrals over General Regions

Transcription

1 Section 6.3 Double Integrls over Generl egions Not ever region is rectngle In the lst two sections we considered the problem of integrting function of two vribles over rectngle. This sitution however is firl unrelistic since most regions in -spce re not rectngles. We need to develop w to clculte integrls over generl regions in the plne. In order to integrte function f(x, ) over n rbitrr region, we shll develop two relted methods dependent upon the structure of the region.. Integrls of tpes nd The first tpe of region we consider is when the boundr of the region consists of two functions of x. We describe below. (i) Suppose we wnt to integrte function f(x, ) over some rbitrr region in the plne which is bounded between two functions g (x) nd g (x) nd between nd b s illustrted below. We cll such region region of tpe. g g b (ii) We shll set up iemnn sum over this region to evlute the definite integrl. First we subdivide the region up into smll rectngles s illustrted where we do not count the rectngles which re not full contined in : g g b (iii) As usul, we tke reference points in ech subrectngle, nd tke iemnn sum over ll the rectngles multipling the re of the rectngle b the function vlue t the reference

2 point in the rectngle. Notice however, tht this cse is different to the cse when we re integrting over rectngles - specificll, the number of rectngles nd the vlues is llowed to tke depend upon x. Specificll, if we choose x i s reference point long column of rectngles, then the vlues re bounded between g (x i ) nd g (x i ), so ll rectngles must be within these bounds. This must be reflected in n iemnn sum we re using to tr to define n integrl. Therefore, we cn define n m f(x, )da f(x i, j ) x i where the second sum is over the rectngles bounded b g (x i ) nd g (x i ). (iv) Tking smller nd smller rectngles, the nswer becomes more exct becuse the rectngles pproximte the region better nd better. Tking them infinitel smll, we get the following iterted integrls: f(x, )da j b g (x) g (x) f(x, )ddx. We cn use similr rgument if the region is bounded b functions in the direction (regions of tpe ), nd thus we get the following importnt w of integrting functions of two vribles over generl regions: esult.. Suppose f(x, ) is continuous function on region. (i) If the region is of Tpe, so bounded bove nd below b functions g (x) nd g (x) nd between x nd x b, then b g (x) f(x, )da f(x, )ddx. g (x) (ii) If the region is of Tpe, so bounded to the left nd right b functions h () nd h () nd between c nd d, then d h () f(x, )da f(x, )dxd. We illustrte with n exmple. c h () Exmple.. Evlute the integrl xda where is the tringle with vertices (, ), (, ) nd (, ).

3 3 First we sketch the region. Observe tht this tringle is bounded between nd the lines x. Therefore it is n integrl of tpe, so xda xdxd x d [( ) ( ) ]d [( + ) ( + )]d d. We could hve lso seen this from the smmetr cross the x-xis! As we observed in the construction of n iterted integrl over n rbitrr region, the order of integrtion is extremel importnt depending upon whether it is region of tpe or tpe. Therefore, it is good ide to hve few generl checkpoints nd useful fcts bout iterted integrls on rbitrr regions to void n sill errors. When setting up n iterted integrl over non-rectngulr region, lws check the following: (i) The limits on the outer integrl must be constnts. (ii) The limits on the inner integrl should be functions of either x for tpe or for tpe. (iii) Integrls re dditive over regions - we cn brek region up into pieces nd dd up the integrl over ech piece to get the integrl over the whole region. (iv) If region is not tpe or tpe, it cn lws be broken up into more integrls which re tpe or nd then we cn use the dditivit of the integrl to evlute it. (v) Much of the time, the difficult is in setting up integrls nd not ctull evluting them. To illustrte the lst point, we do couple of exmples. Exmple.3. Set up n integrl for the function f(x, ) over the following regions.

4 (,) (i) This region is circle of rdius centered t the point (, ), so it hs eqution (x ) + ( ). It cn be relized s either tpe or tpe integrl. elizing it s tpe, we hve nd ( ) x + ( ), so the integrl will be + ( ) ( ) f(x, )dxd. elizing it s tpe region, we hve, x nd (x ) + (x ), so the integrl will be + (x ) f(x, )dxd. (x ) (ii) Consider the following region which is neither tpe or tpe. This region needs to be broken up into different pieces. For x, we cn mke it tpe integrl. Specificll, we hve (x + ). The fourth qudrnt cn lso be mde into tpe region with x nd (x ). The first qudrnt cn be mde into tpe region with nd x

5 5 ( ). Putting these together, we hve (x+) f(x, )da f(x, )ddx+ f(x, )ddx + (x ) We finish with couple more exmples. ( ) Exmple.. Evlute the integrl (3x + )da f(x, )dxd where is qurter of the unit circle centered t the origin in the positive x nd -directions. This region is either tpe or tpe. Using tpe, we hve x nd x, so x (3x + )da (3x + )ddx x 3x + dx (3x x + x )dx ( x ) 3 x 3 + x 3 3 ( ) 5 3 Other exmples to consider in groups: (i) Evlute sin x dxd Note tht we cnnot integrte this function with respect to x - it is function which does not hve n lgebric ntiderivtive, so it looks like we cnnot solve this problem. However, there is nothing to stop us from reversing the order of integrtion provided we cn chnge the bounds s tpe region into tpe region. To do this, we first sketch the region:

6 6 As tpe region, this region will hve the bounds x nd x. Therefore, we hve sin x dxd x x sin (x )dx cos (x ) (ii) Evlute e sin x ddx sin (x ) x dx cos () cos () + ( cos ()) x e ln x dxd The technique we used in the lst exmple is ver useful offering two possible ws to pproch n integrl. As with the lst exmple, we cnnot integrte this function with respect to x, so we shll use the sme technique nd trnsform this from tpe integrl to tpe integrl. First, we sketch the region: As tpe region, this region will hve the bounds ln (x) nd x e. Therefore, we hve e x e ln (x) e ln x dxd x e ln (x) ln x ddx x ln (x) dx e ln (x)x e ln (x) dx xdx x e ( e ). (iii) Evlute x 3 dxd This integrl cn be evluted directl: x 3 x dxd 3 5 d d

7 7 (iv) Evlute x e x ddx This integrl cn be evluted directl: x x e x ddx e x dx xe x dx ex ( e ).