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X
- WE WANT TO USE THE DISK METHOD
TO FIND THE VOLUME OF THE SOLID GENERATED
BY ROTATING THE REGION BOUNDED BY Y = 2.3 RAISE THE POWER OF X.
X = -1, X = 1 ABOUT THE X AXIS.
SO LOOKING AT OUR GRAPH HERE BELOW, HERE'S X = -1.
HERE'S X = 1, AND HERE'S OUR FUNCTION Y = 2.3,
RAISE TO THE POWER OF X.
WE'RE ROTATING ABOUT THE X AXIS HERE.
AND THEREFORE,
THE BOUNDED REGION WOULD BE THIS REGION HERE.
IF WE ROTATE THIS REGION ABOUT THE X AXIS,
IT CREATES THIS SOLID HERE
AND OUR GOAL IS TO FIND THE VOLUME OF THIS SOLID.
WHEN USING THE DISK METHOD OR THE WASHER METHOD
IT'S HELPFUL TO SKETCH A REPRESENTATIVE RECTANGLE
WHERE IF YOU WERE TO ROTATE THE RECTANGLE ABOUT THE X AXIS
IT WOULD GIVE 1 DISK THAT WOULD GIVE US AN APPROXIMATE VOLUME
TO ONE SLICE OF THE SOLID.
AS AN EXAMPLE, IF WE WERE TO USE THIS RECTANGLE HERE
AND ROTATE IT ABOUT THE X AXIS
IT WOULD GIVE US THIS DISK WHICH WOULD GIVE US
THE APPROXIMATE VOLUME OF ONE SLICE OF OUR SOLID.
NOTICE THAT THE RADIUS OF THIS DISK,
A RIGHT CIRCULAR CYLINDER
WOULD BE THIS LENGTH HERE IF WE CALL R OF X, WHICH IN THIS CASE,
WOULD JUST BE THE FUNCTION VALUE 2.3 RAISE TO THE POWER OF X
AND THE WIDTH OF THE RECTANGLE WOULD BE DELTA X,
WHICH IF WE THINK OF THIS AS A RIGHT CIRCULAR CYLINDER
WOULD BE THE HEIGHT OF THE CYLINDER.
SO BECAUSE WE HAVE DELTA X HERE,
WE NEED TO GET WITH RESPECT TO X AND AGAIN,
THE RADIUS WOULD BE THE FUNCTION OF VALUE 2.3,
RAISE TO THE POWER OF X.
REMEMBER IF WE WERE TRYING TO FIND
THE VOLUME OF A RIGHT CIRCULAR CYLINDER
WE WOULD USE THE FORMULA V = PI OR R SQUARED x H.
IF WE TAKE A LOOK AT THE VOLUME FORMULA GIVEN HERE,
WE CAN SEE THIS FORMULA IN OUR INTEGRAL.
WE HAVE THE VOLUME = PI x THE INTEGRAL OF R OF X SQUARED
INTEGRATED WITH RESPECT TO X FROM "A" TO B.
HERE'S THE PI R OF X SQUARED IS THE RADIUS SQUARED
AND DX OR DELTA X REPRESENTS
THE HEIGHT OF THE RIGHT CIRCULAR CYLINDER.
SO WE HAVE ALL THE PIECES WE NEED IN ORDER TO FIND
THE VOLUME OF THIS SOLID USING THE DISK METHOD.
THE VOLUME IS = TO PI x THE INTEGRAL OF R OF X SQUARED
WHICH WOULD BE 2.3 RAISE TO THE POWER OF X SQUARED
INTEGRATED WITH RESPECT TO X FROM -1 TO 1.
NOW LET'S EVALUATE THIS INTEGRAL ON THE NEXT SLIDE.
FIRST NOTICE HOW WE HAVE POWERS TO POWERS HERE.
SO WE'LL MULTIPLY THE EXPONENTS
AND WRITE THIS AS 2.3 RAISE THE POWER OF 2X.
LOOKING AT THE INTEGRATION FORMULA GIVEN HERE BELOW
NOTICE HOW WE'LL HAVE TO PERFORM U SUBSTITUTION
TO INTEGRATE THIS WHERE U WOULD BE = TO 2 X
AND THEREFORE DIFFERENTIALLY U WOULD BE 2X DX.
IF WE DIVIDE BOTH SIDES BY 2 WE CAN SAY 1/2 TO U = DX.
SO FOR THE NEXT STEP
LET'S WRITE THIS IN TERMS OF U WHICH MEANS
WE'LL LEAVE OF THE LIMITS OF INTEGRATION TEMPORARILY.
SO WE'LL HAVE PI x INTEGRAL OF--
THIS WOULD BE 2.3 RAISE TO THE POWER OF U,
BUT DX IS = TO 1/2 TO U, SO WE HAVE AN EXTRA FACTOR OF 1/2.
SO LET'S WRITE THIS AS PI DIVIDED BY 2 TO U,
NOW LET'S FIND THE ANTI-DERIVATIVE
AND WRITE IT BACK IN TERMS OF X.
SO WE HAVE PI/2.
THE ANTI-DERIVATIVE OF 2.3
RAISED TO THE POWER OF U IS GIVEN HERE.
WE'D HAVE 1 DIVIDED BY NATURAL LOG "A"
OR 1 DIVIDED BY NATURAL LOG 2.3 x "A" TO THE U,
WHICH WOULD BE 2.3 RAISE TO THE POWER OF U,
WHICH IS REALLY 2.3 RAISE TO THE POWER OF 2X
AND WE'RE INTEGRATING FROM -1 TO 1.
SO NOW LET£S WRITE THIS
AS PI DIVIDED BY 2 x NATURAL LOG 2.3
AND I WOULD EVALUATE 2.3, RAISE TO THE POWER OF 2X AT (1,-1)
AND THEN FIND THE DIFFERENCE.
SO WE'D HAVE 2.3 RAISE TO THE POWER OF 2 x 1 - 2.3,
RAISE TO THE POWER OF 2 x -1.
SO LET'S WRITE THIS AS PI DIVIDED BY 2 NATURAL LOG 2.3 x
WE'LL LEAVE THIS AS 2.3 SQUARED.
THIS WOULD BE 2.3 TO THE -2 POWER
SO LET'S WRITE THIS AS -1 DIVIDED BY 2.3 SQUARED.
SO THIS WOULD GIVE US THE EXACT VOLUME OF OUR SOLID.
LET'S ALSO GET OUR DECIMAL APPROXIMATION.
SO WE HAVE PI DIVIDED BY 2 NATURAL LOG 2.3
x 2.3 SQUARED - 1 DIVIDED BY 2.3 SQUARED.
IF WE ROUND TO FOUR DECIMAL PLACES THIS WOULD BE
APPROXIMATELY 9.6200
SINCE IN THE 5th DECIMAL PLACE WE HAVE AN 8,
WE ROUND UP MAKING THIS 0.6200.
THIS WAS A FAIRLY INVOLVED INTEGRAL,
SO JUST TO MAKE SURE WE DIDN'T MAKE A MISTAKE
LET'S GO BACK TO THE FIRST SLIDE AND EVALUATE
THIS INTEGRAL HERE USING THE GRAPHING CALCULATOR.
SO WE HAVE PI x THE DIF INTEGRAL SO WE'LL PRESS MATH,
WE'RE LOOKING FOR FUNCTION INTEGRATION
WHICH IS OFF THE SCREEN,
IT LOOKS LIKE IT'S OPTION 9, ENTER.
NEXT ONE ENTER THE INTEGRAND
WHICH IS 2.3 RAISE THE POWER OF 2X,
"COMMA," THE VARIABLE OF INTEGRATION WHICH IS X, "COMMA,"
LOWER LIMIT OF INTEGRATION IS -1, "COMMA,"
UPPER LIMIT OF INTEGRATION IS 1, CLOSED PARENTHESIS AND ENTER.
NOTICE HOW THIS DOES VERIFY OUR RESULTS.
WE HAVE OUR EXACT VOLUME AND THE APPROXIMATE VOLUME.
I HOPE YOU FOUND THIS HELPFUL.