- WELCOME TO A LESSON ON EVALUATING DOUBLE INTEGRALS.
IF WE HAVE A DOUBLE INTEGRAL IN THIS FORM HERE WE INTEGRATE
WITH RESPECTS TO Y FIRST AND THEN INTEGRATE
WITH RESPECTS TO X.
BUT IF IT'S IN THIS FORM WE INTEGRATE WITH RESPECTS TO X
FIRST AND THEN WITH RESPECTS TO Y
AND IF WE'RE INTEGRATING WITH RESPECTS TO X
WE TREAT Y AS A CONSTANT
AND IF WE INTEGRATE WITH RESPECTS Y
WE TREAT X AS A CONSTANT.
LET'S TRY SOME OF THESE.
HERE WE WANT TO EVALUATE THE INTEGRAL
BUT WE ALSO WANT TO SKETCH THE REGION OF INTEGRATION.
SO SINCE WE'RE FIRST INTEGRATING WITH RESPECTS TO Y,
Y WILL BE ON THE CLOSED INTERVAL FROM 1 TO 2
AND X WILL BE ON THE INTERVAL FROM 0 TO 2.
LET'S GO AHEAD AND SKETCH THIS REGION.
X IS BETWEEN 0 AND 2 AND Y IS BETWEEN 1 AND 2.
SO WE WOULD HAVE THE RECTANGULAR REGION THAT LOOKS LIKE THIS.
THIS IS OUR REGION OF INTEGRATION.
X IS ON THE CLOSED INTERVAL FROM 0 TO 2
AND Y IS ON THE CLOSED INTERVAL FROM 1 TO 2.
SO NOW WE'RE GOING TO INTEGRATE WITH RESPECTS TO Y
TREATING X AS A CONSTANT.
SO WE'LL HAVE Y SQUARED/2 MINUS--
3X SQUARED IS A CONSTANT SO WE'D HAVE - 3X SQUARED Y.
LIMITS OF INTEGRATION ARE FROM 1 TO 2.
NOW BECAUSE WE INTEGRATED WITH RESPECTS TO Y
WE'RE GOING TO REPLACE Y WITH THE VALUES OF 2 AND 1.
SO WHEN Y IS 2 WE'LL HAVE 2 SQUARED DIVIDED BY 2
THAT'S 2 - 3X SQUARED x 2 THAT'S 6X SQUARED
AND NOW WE'LL REPLACE Y WITH 1.
WE'LL HAVE 1/2 AND HERE WE'LL HAVE -3X SQUARED.
SO WE'LL HAVE 2 - 1/2 THAT'S 3/2
AND THEN -6X SQUARED - -3X SQUARED
THAT WILL BE - 3X SQUARED
AND NOW WE'RE GOING TO INTEGRATE THIS WITH RESPECTS TO X.
SO WE'LL HAVE 3/2X
THIS WILL BE - 3 x 3 SQUARED/3 OR - X TO THE 3rd.
WE'RE GOING TO EVALUATE THIS AT 0 AND 2.
SO WHEN X IS 2 WE'LL HAVE
3 - THIS IS GOING TO BE 8 AND WHEN X IS ZERO BOTH OF THESE
ARE ZERO SO WE HAVE 3 - 8 SO THIS IS EQUAL TO -5.
REMEMBER A DOUBLE INTEGRAL ONLY GIVES US THE VOLUME
WHEN THE SURFACE IS NONNEGATIVE IN THE REGION
DEFINED HERE IN BLUE.
LET'S TAKE A LOOK AT ANOTHER EXAMPLE.
BECAUSE WE HAVE DY HERE,
WE KNOW THAT Y IS IN THE CLOSED INTERVAL FROM 0 TO X
AND X IS ON THE CLOSED INTERVAL FROM 0 TO 1.
LOOKS LIKE WE HAVE A TRIANGULAR REGION HERE.
X IS BETWEEN 0 AND 1 BUT Y IS BETWEEN 0 AND X.
SO IF WE WERE TO GRAPH Y IS LESS THAN OR EQUAL TO X
THAT'S A LINE WITH A SLOPE OF 1
PASSING THROUGH THE ORIGIN AND SINCE THE MAXIMUM VALUE OF X
IS 1 THE MAXIMUM VALUE FOR Y WOULD ALSO BE 1
SO IT STOPS HERE.
SO OUR REGION OF INTEGRATION IS WHEN X IS IN THE INTERVAL
FROM 0 TO 1 AND Y IS IN THE INTERVAL FROM 0 TO X.
SO IT'S THIS TRIANGULAR REGION HERE.
SO NOW WE'LL INTEGRATE WITH RESPECTS TO Y TREATING X
AS A CONSTANT.
SO WE'LL HAVE THE INTEGRAL FROM 0 TO 1 OF
JUST Y x THE QUANTITY 1 - X SQUARED TO THE 1/2 POWER.
LIMITS OF INTEGRATION ARE FROM 0 TO X
AND AGAIN BECAUSE WE INTEGRATED WITH RESPECTS TO Y
WE'RE REPLACING Y WITH X AND 0.
SO WHEN WE REPLACE Y WITH X
WE'RE GOING TO HAVE X x THE QUANTITY 1 - X SQUARED
TO THE 1/2 AND THEN WHEN Y IS 0 THIS PRODUCT WOULD BE 0.
SO NOW WE JUST NEED TO INTEGRATE WITH RESPECTS TO X.
WELL THIS IS GOING TO REQUIRE U SUBSTITUTION,
U IS GOING TO BE EQUAL TO 1 - X SQUARED.
SO DU IS GOING TO BE EQUAL TO -XDX.
NOTICE OUR INTEGRAND DOES CONTAIN XDX.
SO I'M GOING TO GO AHEAD AND DIVIDE THIS BY -2.
LET'S GO AHEAD AND REWRITE THIS IN TERMS OF U.
XDX IS REPLACED WITH -1/2 DU AND ALL THIS IS U
SO WE HAVE U TO THE 1/2.
SO WE'LL INTEGRATE THIS WE'RE GOING TO HAVE -1/2 x U
TO THE 3/2 DIVIDED BY 3/2
BUT DIVIDING BY 3/2 IS THE SAME AS MULTIPLYING BY 2/3
AND THEN U IS ACTUALLY 1 - X SQUARED.
SO WE NEED TO EVALUATE THIS AT OUR LIMITS OF INTEGRATION.
BEFORE WE DO THAT, WE CAN SEE THIS SIMPLIFIES NICELY.
SO WE HAVE -1/3 x THIS QUANTITY.
LET'S GO AHEAD AND EVALUATE THIS ON THE NEXT SLIDE.
OKAY, SO WHEN X IS 1 WE'RE GOING TO HAVE 1 - 1 HERE
SO THAT'S GOING TO BE ZERO AND THEN WHEN X IS 0
WE'LL HAVE 1 TO THE 3/2 THAT'S JUST GOING TO GIVE US 1.
SO WE HAVE -1/3 x -1 SO THIS GIVES US A 1/3
AND LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE EXAMPLE.
HERE WE'RE GOING TO INTEGRATE WITH RESPECTS TO X FIRST
SO WE'LL HAVE X TO THE 4th/4
AND OUR LIMITS OF INTEGRATION ARE 4 COSINE Y AND 0.
WE'VE SEEN THIS A COUPLE OF TIMES BEFORE.
NOTICE IF WE INTEGRATE WITH RESPECTS TO X
IF THE LIMITS OF INTEGRATION ARE IN TERMS OF FUNCTIONS
THEY MUST BE IN TERMS OF Y.
SO WE'RE GOING TO REPLACE X WITH 4 COSINE Y.
WELL 4 TO THE 4th DIVIDED BY 4 IS GOING TO BE 4 TO THE 3rd
WHICH WILL GIVE US 64,
AND THEN WE HAVE COSINE TO THE 4th Y
AND THEN WHEN X IS ZERO THIS WOULD BE ZERO AGAIN
AND NOW WE NEED TO INTEGRATE THIS.
LET'S GO AHEAD AND FACTOR OUT THE 64.
TO INTEGRATE THIS WE COULD USE A POWER REDUCING FORMULA
BUT WE COULD ALSO RECOGNIZE THAT SINCE THE LIMITS OF INTEGRATION
ARE FROM 0 TO PI/2 AND WE HAVE AN EVEN POWER OF COSINE
WE COULD APPLY WALLIS'S FORMULA.
WHICH TELLS US IF WE HAVE COSINE TO THE POWER OF N
WHERE N IS EVEN AND OUR LIMITS OF INTEGRATION
ARE FROM 0 TO PI/2 THIS IS GOING TO EQUAL 1/2 x 3/4 x 4/5
ALL THE WAY TO N - 1/N x PI/2.
SO FOR OUR INTEGRAL WE HAVE N = 4
SO WE'RE GOING TO HAVE 64 x 1/2 x 3/4
AND NOW WE'RE GOING TO STOP BECAUSE IF N IS 4
THE LAST FRACTION WOULD BE N - 1/N OR 3/4 AND THEN x PI/2,
AND THIS SIMPLIFIES VERY NICELY
BECAUSE 2 x 4 x 2 THAT WOULD BE 16
AND THERE ARE FOUR FACTORS OF 16 IN 64.
SO WE'RE LEFT WITH 4 x 3 x PI OR 12PI
AND SO WALLIS'S FORMULA SAVES US QUITE A BIT OF TIME
IN THIS PROBLEM RATHER THAN USING THE POWER
REDUCING FORMULAS.
I HOPE YOU FOUND THESE EXAMPLES HELPFUL.
THANK YOU FOR WATCHING.