- WELCOME TO PART 1 OF DOUBLE INTEGRALS AND VOLUME
OVER A GENERAL REGION.
BY GENERAL REGION, WE'RE TALKING ABOUT REGIONS
THAT ARE NOT RECTANGULAR.
WE'VE ALREADY DISCUSSED THE LIMIT DEFINITION
OF A DOUBLE INTEGRAL TO DETERMINE THE VOLUME
UNDER A SURFACE.
SO LET'S GO AHEAD AND TAKE A LOOK
AT HOW WE'RE GOING TO DEAL WITH REGIONS
THAT ARE NOT RECTANGULAR.
IF THE REGION IS BOUNDED ABOVE AND BELOW BY A FUNCTION OF X
THEN WE'LL HAVE TO INTEGRATE WITH RESPECTS TO Y FIRST
WITH THE LIMITS OF INTEGRATION WITH RESPECTS TO Y WILL BE
G OF X TO H OF X,
AND IF OUR REGION IS BOUNDED TO THE LEFT AND RIGHT
BY FUNCTIONS OF Y WE'LL FIRST INTEGRATE WITH RESPECTS TO X
AND OUR LIMITS OF INTEGRATION WILL BE G OF Y AND H OF Y.
LET'S GO AND TAKE A LOOK AT OUR EXAMPLES.
WE WANT TO DETERMINE THE VOLUME UNDER THE PLANE Z = X + Y
AND OUR REGION OF INTEGRATION IS DEFINED HERE
WHERE X IS IN THE CLOSED INTERVAL FROM 0 TO Y/2
AND Y IS IN THE CLOSED INTERVAL FROM 0 TO 6.
SO IT'S IMPORTANT FOR THESE TYPES OF PROBLEMS
THAT WE GRAPH OUR REGION OF INTEGRATION.
SO LETS TAKE A CLOSE LOOK AT THIS INTERVAL HERE.
WE KNOW X HAS TO BE GREATER THAN OR EQUAL TO ZERO
BUT X ALSO HAS TO BE LESS THAN OR EQUAL TO Y/2.
SO IF WE KNOW X IS GREATER THAN OR EQUAL TO ZERO
WE'RE TALKING ABOUT THE X AXIS HERE
BUT LET'S TAKE A LOOK AT THIS INEQUALITY HERE.
IF WE WERE TO SOLVE THIS EQUATION FOR Y
BY MULTIPLYING BOTH SIDES BY 2
WE WOULD HAVE 2X LESS THAN OR EQUAL TO Y.
LET'S GO AHEAD AND FLIP THIS AROUND.
THIS IS THE SAME AS Y IS GREATER THAN OR EQUAL TO 2X.
SO IF WE WERE TO GRAPH THIS INEQUALITY
WE'D HAVE A SOLID LINE HERE BUT WE'D SHADE ABOVE THE LINE.
SO THAT TELLS US THE INTERVAL FOR X
IS GOING TO BE STARTING HERE WHERE X IS ZERO
AND THEN MOVING TO THE RIGHT
UNTIL WE APPROACH THIS LINE HERE.
SO IT'S A COMMON MISTAKE FROM THIS INTERVAL HERE
TO THINK IT'S GOING TO BE BELOW THIS LINE
BUT IT'S ACTUALLY ABOVE THIS LINE
AND TO THE RIGHT OF THE Y AXIS
AND THEN Y IS ON THE CLOSED INTERVAL FROM 0 TO 6.
SO IT'S GOING TO BE THIS TRIANGULAR REGION
HERE ABOVE THE LINE Y = 2X.
THIS WOULD BE OUR REGION OF INTEGRATION.
NOW THAT WE HAVE OUR REGION OF INTEGRATION
LET'S GO AHEAD AND SET UP OUR DOUBLE INTEGRAL
AND SINCE THE INTERVAL FOR X IS GIVEN IN TERMS OF Y
WE'LL HAVE TO INTEGRATE WITH RESPECTS TO X FIRST.
SO F OF XY WOULD BE X + Y.
INTEGRATE WITH RESPECTS TO X FIRST AND THEN Y.
SO FOR THE LIMITS OF INTEGRATION FOR X WE'LL HAVE FROM 0 TO Y/2
AND THEN FOR Y IT'LL BE FROM 0 TO 6.
SINCE WE MENTIONED THIS BEFORE IF WE INTEGRATE WITH RESPECTS
TO X THE LIMITS OF INTEGRATION WOULD BE IN TERMS OF Y.
SO TREATING Y AS A CONSTANT
WE'RE GOING TO HAVE X SQUARED/2 + YX.
WE NEED TO EVALUATE THIS.
BECAUSE WE INTEGRATED WITH RESPECTS TO X,
WE'RE REPLACING X WITH Y/2.
SO HERE WE'RE GOING TO HAVE 1/2 x Y/2 SQUARED + Y x Y/2 -
AND THEN WHEN X IS ZERO THESE WILL BOTH BE ZERO.
SO LET'S GO AHEAD AND CLEAN THIS UP.
HERE WE'RE GOING TO HAVE Y SQUARED/8 + Y SQUARED/2.
SO IF WE GET A COMMON DENOMINATOR OF 8,
THIS IS GOING TO BE 5Y SQUARED/8.
LET'S GO AHEAD AND TAKE THIS ON TO THE NEXT SLIDE.
NOW WE'LL INTEGRATE WITH RESPECTS TO Y.
WE'LL HAVE 5/8 x Y TO THE 3rd/3.
WE'LL EVALUATE THIS AT 6 AND 0.
SO WE'RE GOING TO HAVE 5/24 AND THEN WHEN Y IS 6
WE'LL HAVE 6 CUBED AND THEN WHEN Y IS 0 WE'LL HAVE 0 CUBED.
SO WE HAVE 5/24 x 6 TO THE 3rd.
I'VE ALREADY WORKED THIS OUT.
IT COMES OUT TO 45 AND BECAUSE THIS REPRESENTS
THE VOLUME UNDER THAT PLANE THIS WOULD BE 45 CUBIC UNITS.
LET'S TAKE A LOOK AT A GRAPH OF THIS.
IN RED, YOU CAN SEE THE REGION OF INTEGRATION
AND I'VE BOUNDED THAT REGION BY THESE YELLOW PLANES.
SO WHAT WE JUST FOUND WAS THE VOLUME
UNDER THE PURPLE PLANE BOUNDED BY THESE YELLOW PLANES
AND THAT WAS 45 CUBIC UNITS.
LET'S TAKE A LOOK AT ANOTHER EXAMPLE.
HERE WE WANT TO DETERMINE THE VOLUME OF THE GIVEN FUNCTION
IN THE REGION WHERE X IS ON THE CLOSED INTERVAL
FROM 0 TO 1 AND Y IS ON THE CLOSED INTERVAL
FROM X SQUARED TO X.
LET'S TAKE A CLOSER LOOK AT THIS INTERVAL HERE.
THIS COMPOUNDED INEQUALITY CAN BE WRITTEN TWO WAYS.
STARTING FROM THE CENTER MOVING TO THE RIGHT
WE HAVE Y IS LESS THAN OR EQUAL TO X
AND THEN FROM THE CENTER TO THE LEFT
WE HAVE Y IS GREATER THAN OR EQUAL TO X SQUARED.
SO Y WOULD BE BELOW THE LINE
Y = X AND ABOVE THE PARABOLA Y = X SQUARED
AND X IS ON THE CLOSED INTERVAL FROM 0 TO 1
SO OUR REGION OF INTEGRATION IS THIS REGION HERE
BETWEEN THE PARABOLA AND THE LINE
AND AGAIN, THIS WAS THE POINT (1, 1).
NOTICE THIS REGION IS BOUNDED ABOVE AND BELOW
AS WELL AS TO THE LEFT AND RIGHT BY THESE TWO FUNCTIONS.
BUT SINCE WE HAVE Y IN TERMS OF X
WE'LL INTEGRATE WITH RESPECTS TO Y FIRST THIS TIME.
SO OUR FUNCTION IS XY.
WE'LL HAVE DY FIRST AND THEN DX.
SO Y EXPRESSED IN TERMS OF X WILL BE FROM X SQUARED TO X
AND X WILL BE FROM 0 TO 1.
SO NOW WE'LL INTEGRATE WITH RESPECTS TO Y
TREATING X AS A CONSTANT.
SO WE'LL HAVE XY SQUARED/2 IN OUR LIMITS OF INTEGRATION.
BECAUSE WE INTEGRATED WITH RESPECTS TO Y,
WE'RE REPLACING Y WITH X AND X SQUARED.
SO WHEN Y = X WE'LL HAVE X CUBED/2 - AND THEN
WHEN Y IS X SQUARED WE'LL HAVE X TO THE 5th/2.
LET'S GO AHEAD AND FINISH THIS ON THE NEXT SLIDE.
SO WE'LL HAVE 1/2 x X TO THE 4th/4 - 1/2X TO THE 6th/6
WE'LL EVALUATE THIS AT 1 AND 0.
LET'S CLEAN IT UP FIRST THOUGH.
WE'LL HAVE 1/8X TO THE 4th - 1/12X TO THE 6th.
SO WHEN X IS 1 WE'LL HAVE 1/8 - 1/12 AND THEN WHEN X IS 0
WE'LL HAVE 0 - 0.
OUR COMMON DENOMINATOR WOULD BE 24.
SO WE HAVE 3/24 - 2/24 THAT WILL BE 1/24
AND THIS WOULD BE THE VOLUME UNDER THAT SURFACE
OVER THE DEFINED REGION.
LET'S GO AND TAKE A LOOK AT THIS ONE GRAPHICALLY AS WELL.
YOU CAN SEE THIS OUTLINED IN RED AS OUR REGION OF INTEGRATION.
SO THE VALUE THAT WE FOUND WAS THE VOLUME OF THIS REGION HERE
BELOW THE PURPLE SURFACE BOUNDED BY THIS REGION IN YELLOW
AND IT WAS 1/24 OF A CUBIC UNIT.
THAT WILL DO IT FOR PART 1.
WE'LL TAKE A LOOK AT SOME MORE EXAMPLES IN PART 2.