- WELCOME TO PART 2 OF DOUBLE INTEGRALS
AND VOLUME OVER A GENERAL REGION.
REMEMBER A GENERAL REGION MEANS A REGION
WILL NOT BE RECTANGULAR.
LET'S START WITH A QUICK REVIEW
OF HOW WE'RE GOING TO SET THESE INTEGRALS UP.
THE REGION IS BOUNDED ABOVE AND BELOW BY A FUNCTION OF X.
WE'RE GOING INTEGRATE WITH RESPECTS TO Y FIRST
USING G OF X AND H OF X AS OUR LIMITS OF INTEGRATION
IN TERMS OF Y AND IF THE REGION IS BOUNDED TO THE LEFT AND RIGHT
BY A FUNCTION OF Y WE'LL INTEGRATE WITH RESPECTS TO X
FIRST THE LIMITS OF INTEGRATION BEING G OF Y AND H OF Y.
SO LET'S GO AHEAD AND TAKE A LOOK AT OUR FIRST EXAMPLE.
WE WANT TO DETERMINE THE VOLUME OF THE GIVEN FUNCTION
IN THE REGION BOUNDED BY X GREATER THAN OR EQUAL TO ZERO,
Y = X - 1 AND Y = 1/2X.
IF WE LOOK AT THE GRAPH DOWN HERE THIS SEGMENT HERE
WOULD COME FROM X = 0.
THIS SEGMENT HERE WOULD BE ON THE LINE Y = X - 1
AND THIS SEGMENT HERE WOULD BE ON THE LINE Y = 1/2X.
SO WE CAN SEE THE BOUNDED REGION IS THIS TRIANGULAR REGION
WHICH REPRESENTS OUR REGION OF INTEGRATION.
SO LOOKING AT THIS REGION HERE WE NEED TO DECIDE
WHETHER WE'RE GOING TO INTEGRATE WITH RESPECTS TO Y FIRST
OR WITH RESPECTS TO X FIRST.
THIS REGION IS BOUNDED ABOVE AND BELOW BY THESE TWO FUNCTIONS
IN TERMS OF X.
IF WE TAKE A LOOK FROM THE LEFT TO THE RIGHT
IT ACTUALLY CHANGES.
FROM THE LEFT TO RIGHT IT'S BOUNDED BY THESE TWO LINES
AND THEN ONCE WE DROP BELOW THE X AXIS IT'S BOUNDED BY THIS LINE
AND X = 0.
SO SINCE THIS ENTIRE REGION IS BOUNDED VERTICALLY
BY TWO FUNCTIONS IN TERMS OF X
WE'LL INTEGRATE WITH RESPECTS TO Y FIRST
AND THEN WITH RESPECTS TO X.
SO LET'S GO AHEAD AND SET THIS UP.
OUR FUNCTION IS Y SQUARED.
BECAUSE IT'S BOUNDED ABOVE AND BELOW BY TWO FUNCTIONS
WE'LL INTEGRATE WITH RESPECTS TO Y FIRST
AND THEN WITH RESPECTS TO X.
SO LIMITS OF INTEGRATION WITH RESPECTS TO Y
ARE GOING TO BE X - 1 TO 1/2X AND REMEMBER IF WE --
WITH RESPECTS TO Y THESE LIMITS OF INTEGRATION
MUST BE IN TERMS OF X
AND THEN FOR X THE LIMITS OF INTEGRATION WILL BE FROM ZERO
ALL THE WAY OUT TO THIS POINT OF INTERSECTION HERE.
THIS POINT IS (2, 1).
SO X WILL BE FROM 0 TO 2 AND NOW WE'LL INTEGRATE
WITH RESPECTS TO Y.
THAT'S JUST GOING TO GIVE US Y TO THE 3rd/3.
NOW WE'LL REPLACE Y WITH 1/2X.
WE'LL HAVE 1/3 x 1/2X CUBED - WE'LL HAVE X - 1
TO THE 3rd DIVIDED BY 3.
THIS WILL BE 1/3 x 1/8X TO THE 3rd.
THIS WILL BE 1/24X CUBED.
WE'LL LEAVE THIS AS - 1/3 x THE QUANTITY X - 1 TO THE 3rd.
WE'LL INTEGRATE ALL THIS WITH RESPECTS TO X ON THE NEXT SLIDE.
SO HERE WE'RE GOING TO HAVE 1/24 x X TO THE 4th/4 - 1/3.
HERE WE'RE GOING TO HAVE X - 1 TO THE 4th/4
WHEN YOU EVALUATE THIS AT 2 AND 0.
LET'S GO AHEAD AND CLEAN THIS UP ONE MORE TIME
AND THEN WE'LL PERFORM THE SUBSTITUTION.
THIS WILL GIVE US
1/96X TO THE 4th - THIS WILL BE 1/12 x THE QUANTITY X - 1
TO THE 4th.
SO WE'LL HAVE 1/96, 2 TO THE 4th - 1/12
WHEN X IS 2 WE'RE GOING TO HAVE
1 TO THE 4th HERE - AND THEN WHEN X IS 0 THIS WILL BE 0.
THIS WILL BE 1/12 x -1 TO THE 4th. SO THIS ENDS UP BEING 1/6.
THIS WILL BE - 1/12.
OVER HERE, WE'RE GOING TO HAVE - -1/12 THAT WILL BE + 1/12.
SO FINALLY, WE HAVE 1/6.
SO THE VOLUME OVER THAT TRIANGULAR REGION
WOULD BE 1/6 AS A CUBIC UNIT.
LET'S TAKE A LOOK AT THE GRAPH.
IN PURPLE, WE SEE THE SURFACE
AND IN RED WE HAVE THE TRIANGULAR REGION
IN THE XY PLANE.
SO THAT 1/24 REPRESENTS THE VOLUME
UNDER THE SURFACE ABOVE THE XY PLANE IN THIS RED REGION.
LET'S GO AND TAKE A LOOK AT ANOTHER EXAMPLE.
HERE WE HAVE OUR FUNCTION
AND THE REGION IS BOUNDED BY Y = X CUBED
AND Y = THE SQUARE ROOT OF X.
SO THE MORE WE KNOW ABOUT THESE FUNCTIONS
THE EASIER IT SHOULD BE.
TAKING A LOOK AT THIS GRAPH HERE BELOW THIS
WILL BE THE GRAPH OF Y = X CUBED
AND THIS PIECE HERE WOULD BE THE GRAPH
OF Y = THE SQUARE ROOT OF X.
NOTICE THE POINT OF INTERSECTION WOULD BE THE POINT (1, 1).
SO WHEN X = 1 Y = 1 FOR BOTH OF THESE EQUATIONS.
SO THE REGION OF INTEGRATION IS THIS REGION HERE
AND TO SET THIS DOUBLE INTEGRAL UP WE NEED TO DECIDE
WHETHER WE'RE GOING TO INTEGRATE WITH RESPECTS TO Y FIRST
OR WITH RESPECTS TO X FIRST.
NOTICE THAT IT'S BOUNDED BY THE TWO FUNCTIONS ABOVE AND BELOW
AS WELL AS TO THE LEFT AND TO THE RIGHT.
BUT BECAUSE THESE FUNCTIONS ARE ALREADY SOLVED FOR Y
WE'RE GOING TO INTEGRATE WITH RESPECTS TO Y FIRST
AND THEN WITH RESPECTS TO X.
SO OUR FUNCTION IS 3XY - X CUBED AND WE'LL HAVE DY/DX.
SO LIMITS OF INTEGRATION
ARE GOING TO BE FROM X CUBED TO THE SQUARE ROOT OF X
AND THEN FOR Y IT WILL BE FROM 0 TO 1.
IF WE WANTED TO INTEGRATE WITH RESPECTS TO X FIRST
WE WOULD HAVE TO SOLVE BOTH OF THESE EQUATIONS FOR X
SO THAT WE COULD EXPRESS THE LIMITS OF INTEGRATION
FOR X IN TERMS OF Y.
SO IT'S A LITTLE BIT LESS WORK TO SET IT UP IN THIS ORDER.
SO WE'LL INTEGRATE THIS WITH RESPECTS TO Y
TREATING X AS A CONSTANT.
SO HERE WE'LL HAVE 3XY SQUARED/2 - X CUBED Y.
THE LIMITS OF INTEGRATION ARE X CUBED
AND I'M GOING TO GO AHEAD AND WRITE THIS AS
X TO THE 1/2 POWER.
LET'S SEE WHAT WE GET HERE.
BECAUSE WE INTEGRATED WITH RESPECTS TO Y
WE'RE REPLACING Y WITH X TO THE 1/2 AND X TO THE 3rd
AND THAT'S IMPORTANT TO REMEMBER.
SO WHEN Y = X TO THE 1/2 WE'LL HAVE 3/2X SQUARED - X TO THE 3rd
x X TO THE 1/2 THAT'S X TO THE 7/2 - NOW WE REPLACE Y
WITH X CUBED.
THIS WILL BE X x X TO THE 6th.
SO WE HAVE 3/2X TO THE 7th AND THIS WILL BE X TO THE 3rd x X
TO THE 3rd - X TO THE 6th.
LET'S GO AND CONTINUE THIS ON THE NEXT SLIDE.
LET'S GO AHEAD AND CLEAR THE PARENTHESES HERE.
NOTICE WE DON'T HAVE ANY LIKE TERMS
SO WE'LL INTEGRATE THIS NOW.
WE'LL HAVE 3/2 x X TO THE 3rd/3 - 7/2 + 1 WOULD BE 9/2
SO WE'LL HAVE 2/9X TO THE 9/2 - 3/2X TO THE 8th/8 + X
TO THE 7th/7.
NOW WE NEED TO EVALUATE THIS AT 1 AND 0.
BUT LET'S GO AHEAD AND CLEAN THIS UP BEFORE WE DO THAT.
THIS IS GOING TO BE 1/2X CUBED THE ONLY THING WE CAN DO HERE.
THIS IS GOING TO BE - 3/16X TO THE 8th + 1/7X TO THE 7th.
OKAY, SO WHEN X = 1
WE'LL HAVE 1/2 - 2/9 - 3/16 + 1/7 - AND THEN WHEN X IS 0
ALL OF THESE TERMS WILL BE 0 AND AGAIN, TO SAVE TIME
I'VE ALREADY CALCULATED THIS.
IT DOESN'T COME OUT VERY NICE. IT'S 235/1008.
AGAIN, THIS WOULD BE THE VOLUME
UNDER THE SURFACE IN THE GIVEN REGION.
LET'S TAKE A LOOK AT THE GRAPH OF THIS.
SO OUTLINED IN RED WE CAN SEE THE REGION OF INTEGRATION
AND THEN IN PURPLE WE HAVE THE SURFACE
AND SO THE VOLUME THAT WE FOUND WAS THE VOLUME
IN THIS REGION HERE AND THAT'S ALL FOR THIS VIDEO.
THANK YOU FOR WATCHING.