- WELCOME TO A LESSON ON DOUBLE INTEGRALS AND VOLUME.
THIS LIMIT DEFINITION FOR VOLUME SHOULD LOOK VERY FAMILIAR
TO WHAT WE SAW IN CALC 1 FOR THE LIMIT DEFINITION OF AREA.
HOWEVER, FOR VOLUME WE'RE GOING TO BE SUMMING THE VOLUME
OF RECTANGULAR PRISMS OR BOXES.
TO DETERMINE THE VOLUME OF EACH OF THESE BOXES
WE'RE GOING TO DETERMINE THE AREA OF THE BASE
AND THEN MULTIPLY IT BY THE HEIGHT.
IF WE TAKE A LOOK AT JUST ONE BOX THE AREA OF THE BASE,
DELTA "A" WOULD BE DETERMINED BY MULTIPLYING DELTA X x DELTA Y.
ONCE WE HAVE THE AREA OF THE BASE WE PICK A POINT XY
IN THIS REGION PLUG IT INTO THE FUNCTION
AND THIS GIVES US THE HEIGHT OF THE BOX.
SO THE AREA OF THIS BASE x THE HEIGHT OF THE BOX GIVES US
THE VOLUME OF EACH OF THESE BOXES AND OF COURSE
AS WE KNOW FROM CALC 1 IF WE APPROXIMATE THE VOLUME
UNDER THIS SURFACE WITH MORE AND MORE BOXES WE WILL APPROACH
THE VOLUME UNDERNEATH THE SURFACE.
SO THIS INFINITE SUM HERE
REPRESENTS THE AREA OF THOSE BOXES.
AGAIN THIS IS THE AREA OF THE BASE,
THIS WOULD BE THE HEIGHT AND AS THE NUMBER OF BOXES
APPROACHES INFINITY WE'RE APPROACHING
THE VOLUME UNDER THE SURFACE
AND THIS IS DEFINED BY THE DOUBLE INTEGRAL
OVER THE SPECIFIED REGION IN THE XY PLANE OF F OF XY
AND NOTICE HERE IT'S SHOWING WE'RE DIFFERENTIATING
WITH RESPECTS TO "A."
WELL DIFFERENTIAL "A" CAN BE WRITTEN AS DX/DY OR DY/DX
BECAUSE EITHER WAY IT REPRESENTS THE AREA OF THE BASE
OF EACH BOX.
SO WHAT I MEAN BY THAT IS,
HERE'S THE FORM OF A DOUBLE INTEGRAL THAT WE NORMALLY SEE
WHERE R IS A REGION DEFINED BY X AND Y.
SO FIRST WE NEED TO DECIDE
WHETHER WE WANT TO INTEGRATE WITH RESPECTS TO X FIRST
AND THEN Y OR WITH THE RESPECTS TO Y FIRST AND THEN X.
WE CAN USE EITHER ORDER
BUT OF COURSE IT WILL CHANGE HOW WE ARRANGE
OUR LIMITS OF INTEGRATION.
LET'S TAKE A LOOK AT SOME EXAMPLES.
HERE WE WANT TO DETERMINE THE VOLUME OF THE GIVEN SURFACE
ON THE CLOSED INTERVAL FROM 0 TO 2 FOR X
AND THE CLOSED INTERVAL OF 0 TO 3 FOR Y.
SO LOOKING AT THE XY PLANE
THIS WOULD BE OUR REGION OF INTEGRATION
AND WE ALSO NEED TO MAKE SURE
THAT THIS IS IN THE FORM OF F OF XY.
SO LET'S GO AHEAD AND REWRITE THIS.
F OF XY WOULD BE EQUAL TO 12 - X SQUARED - 1/2Y SQUARED.
NOW WE NEED TO DECIDE WHETHER WE WANT TO INTEGRATE
WITH RESPECTS TO X FIRST OR WITH RESPECTS TO Y.
LET'S GO AHEAD AND SET THIS UP
WHERE WE INTEGRATE WITH RESPECTS TO X FIRST AND THEN Y.
AGAIN, WE'LL FIRST INTEGRATE WITH RESPECTS TO X
AND THEN WITH RESPECTS TO Y.
SO LIMITS OF INTEGRATION FOR X WILL BE FROM 0 TO 2
AND FOR Y IT WILL BE FROM 0 TO 3.
SO WE'LL FIRST INTEGRATE WITH RESPECTS TO X TREATING Y
AS A CONSTANT.
SO WE'LL HAVE 12X - X TO THE 3rd/3 - 1/2Y SQUARED X.
REMEMBER Y IS THE CONSTANT.
WE'LL EVALUATE THIS AT THE LIMITS OF INTEGRATION
AND THEN WE'LL INTEGRATE WITH RESPECTS TO Y
AND BECAUSE WE INTEGRATED WITH RESPECTS TO X
WE'RE GOING TO REPLACE X WITH 2 AND 0.
SO WHEN X IS 2 THIS WILL BE 24 - 8/3 - Y SQUARED
AND WHEN X IS 0 EACH OF THESE TERMS WILL BE 0.
LET'S GO AHEAD AND FINISH THIS ON THE NEXT SLIDE.
LET'S GO AHEAD AND SUBTRACT THESE.
WE'RE GOING TO HAVE 64/3 - Y SQUARED
AND NOW WE'LL INTEGRATE WITH RESPECTS TO Y.
WE'RE GOING TO HAVE 64/3Y - Y TO THE 3rd/3.
NOW WE'LL REPLACE Y WITH 3 AND THEN Y WITH ZERO
AND THESE WILL BOTH BE ZERO.
SO WE'RE GOING TO HAVE 64 HERE
AND THIS WILL BE - 9 WHICH GIVES US 55.
SINCE THIS REPRESENTS THE VOLUME UNDER THAT SURFACE
ON THE GIVEN REGION THIS WILL BE CUBIC UNITS.
LET'S TAKE A LOOK AT THIS IN THREE DIMENSIONS.
SO IN YELLOW WE SEE THE SURFACE
AND THEN THESE PLANES REPRESENT THE REGION.
SO THE AMOUNT OF VOLUME UNDER THE YELLOW SURFACE
ABOVE THE XY PLANE WOULD BE 55 CUBIC UNITS.
LET'S GO AND TAKE A LOOK AT ANOTHER EXAMPLE.
THERE ARE SEVERAL WAYS TO DEFINE A REGION.
IF IT'S DEFINED LIKE THIS WE ALWAYS ASSUME
THAT THIS INTERVAL IS FOR X AND THIS INTERVAL IS FOR Y.
SO IF X IS ON THE CLOSED INTERVAL FROM 1 TO 2
AND Y IS ON THE CLOSED INTERVAL FROM 0 TO 3
OUR REGION OF INTEGRATION IN THE XY PLANE
WOULD BE THIS REGION HERE.
IT'S A GOOD HABIT TO SKETCH THIS
BECAUSE IF IT'S NOT A RECTANGULAR REGION
THIS SKETCH HERE BECOMES VERY IMPORTANT.
SO IN THIS GRAPH WE SEE THE SURFACE
GRAPHED OVER THIS REGION.
SO WE'RE TRYING TO FIND THE VOLUME UNDER THIS SURFACE.
LET'S GO AHEAD AND INTEGRATE
WITH RESPECTS TO Y FIRST THIS TIME.
F OF XY IS XY SQUARED AND WE'LL INTEGRATE
WITH RESPECTS TO Y FIRST AND THEN X.
SO THE INTERVAL FOR Y IS FROM 0 TO 3.
THE INTERVAL FOR X IS FROM 1 TO 2.
SO WE'LL FIRST INTEGRATE WITH RESPECTS TO Y.
TREATING X AS A CONSTANT WE'LL HAVE XY TO THE 3rd/3
WE EVALUATE THIS IN OUR LIMITS OF INTEGRATION
AND THEN INTEGRATE WITH RESPECTS TO X
AND AGAIN, BECAUSE WE INTEGRATED WITH RESPECTS TO Y FIRST
WE'RE REPLACING Y WITH 3 AND Y WITH ZERO.
SO WHEN Y IS 3 THIS IS GOING TO END UP BEING 9X
AND THEN WHEN Y IS ZERO IT'LL BE ZERO.
SO NOW WE'LL INTEGRATE WITH RESPECTS TO X
THAT'LL GIVE US 9 x X SQUARED/2.
SO WHEN X IS 2 WE'RE GOING TO HAVE 4 DIVIDED BY 2
THAT'S 2 x 9 THAT'S 18 AND THEN WHEN X IS 1
WE'LL HAVE 9 x 1/2 OR 9/2.
WELL 18 - 9/2 THAT WOULD BE 36/2 - 9/2.
THAT WILL GIVE US 27/2.
AND AGAIN, THIS REPRESENTS THE VOLUME
UNDER THE SURFACE IN THIS REGION IN THE XY PLANE.
LET'S TAKE A LOOK AT THIS IN 3D AS WELL.
IN YELLOW, WE SEE THE SURFACE
AND THE REGION FOR THIS PROBLEM IS THIS REGION IN HERE
SO WE'RE LOOKING FOR THE VOLUME UNDER THE YELLOW SURFACE
IN THIS REGION LOOKING FOR THIS VOLUME IN HERE.
LET'S TAKE A LOOK AT ONE MORE EXAMPLE.
ON THIS EXAMPLE, ARE REGION IS THE CLOSED INTERVAL
FROM 0 TO 1 FOR X AND FROM 0 TO PI/2 FOR Y.
SO AGAIN WE HAVE A RECTANGULAR REGION
THAT LOOKS SOMETHING LIKE THIS.
SO THIS WOULD BE OUR REGION OF INTEGRATION.
NOW SOMETIMES WHEN SETTING UP A DOUBLE INTEGRAL
THE ORDER OF INTEGRATION DOES MATTER BASED UPON OUR INTEGRAND.
MEANING FOR THIS ONE IT'S GOING TO BE MUCH EASIER
IF WE INTEGRATE WITH RESPECTS TO Y FIRST AND THEN X
BECAUSE IF WE INTEGRATE WITH RESPECTS TO Y FIRST
WE CAN TREAT THIS X HERE AS A CONSTANT
OTHERWISE IT WILL BE MUCH MORE CHALLENGING TO INTEGRATE.
SO IF WE INTEGRATE WITH RESPECTS TO Y FIRST
THE LIMITS OF INTEGRATION FOR Y WOULD BE FROM 0 TO PI/2
AND FOR X IT'LL BE FROM 0 TO 1.
NOW EVEN THOUGH WE'RE TREATING X AS A CONSTANT
WE DO STILL HAVE TO PERFORM U SUBSTITUTION
BECAUSE WE HAVE THE COSINE OF X x Y.
SO IF WE LET U = XY DIFFERENTIAL U WITH RESPECTS TO Y
WOULD BE XDY AND THIS IS ACTUALLY GOOD NEWS
BECAUSE THE INTEGRAND DOES CONTAIN XDY.
SO THIS JUST BECOMES COSINE U
AND THE ANTIDERIVATIVE OF COSINE U IS SINE U.
SO WE'LL HAVE SINE U IS ACTUALLY XY.
OF COURSE WE HAVE TO EVALUATE THIS
AT OUR LIMITS OF INTEGRATION
AND THEN INTEGRATE WITH RESPECTS TO X
AND BECAUSE WE INTEGRATED WITH RESPECTS TO Y
WE'RE REPLACING Y WITH PI/2 AND 0.
SO WHEN Y IS PI/2 WE'LL HAVE SINE PIX/2
AND THEN WHEN Y IS 0 WE'LL HAVE THE SINE OF 0
WHICH IS JUST 0.
LET'S GO AHEAD AND FINISH THIS ON THE NEXT SLIDE,
WE DO HAVE TO PERFORM U SUBSTITUTION AGAIN.
U IS GOING TO BE PIX/2.
SO DIFFERENTIAL U IS GOING TO BE
PI/2 DX WHICH MEANS DX = 2/PI DU.
SO WE'LL HAVE AN EXTRA FACTOR OF 2 DIVIDED BY PI
AND THE ANTIDERIVATIVE OF SINE U IS GOING TO BE NEGATIVE COSINE U
SO WE'LL HAVE NEGATIVE COSINE PIX/2.
LET'S SEE WHAT THIS GIVES US.
WHEN X = 1 WE HAVE NEGATIVE COSINE OF PI/2.
WELL COSINE PI/2 IS 0 AND THEN THE COSINE OF 0 WOULD BE 1
BUT NOTICE WE HAVE A NEGATIVE SIGN THERE SO IT'S -1.
SO WE END UP HAVING PI/2 x 1 WHICH IS JUST PI/2.
SO THIS WOULD BE THE VOLUME UNDER THIS SURFACE
OVER THIS REGION IN THE XY PLANE.
THAT WILL DO IT FOR THIS VIDEO.