- WE WANT TO USE THE DISK METHOD TO FIND THE VOLUME
OF THE SOLID GENERATED
BY ROTATING THE REGION BOUNDED BY Y = SINE SQUARED X
AND Y = 0 ON THE CLOSED INTERVAL FROM 0 TO PI ABOUT THE X AXIS.
LOOKING AT THE GRAPH HERE ON THE LEFT,
THIS IS THE BOUNDED REGION
THE BLUE GRAPH IS OUR FUNCTION Y = SINE SQUARED X.
WE'RE ROTATING THIS REGION ABOUT THE X AXIS
AND OUR GOAL IS TO FIND THE VOLUME OF THIS SOLID HERE.
WHEN USING THE DISK METHOD
IT'S HELPFUL TO SKETCH A REPRESENTATIVE RECTANGULAR
WHERE IF YOU WERE TO ROTATE THE RECTANGLE IT WOULD BE
THE APPROXIMATE VOLUME OF ONE SLICE OF OUR SOLID.
AS AN EXAMPLE, IF WE USE THIS RECTANGLE HERE
AND ROTATE IT ABOUT THE X AXIS IT WOULD GIVE US A DISK
THAT WOULD BE THE APPROXIMATE VOLUME
OF ONE SLICE OF OUR SOLID MAYBE THIS SLICE HERE.
TO FIND THE VOLUME USING THE DISK METHOD
WE USE THIS FORMULA HERE
WHERE THE VOLUME IS EQUAL TO PI x INTEGRAL OF R OF X SQUARED
INTEGRATED WITH RESPECT TO X FROM "A" TO B.
WHERE R OF X WOULD BE THE RADIUS OF ANY OF THE DISKS.
SO LOOKING AT THIS RECTANGLE AGAIN,
NOTICE HOW THIS LENGTH HERE
WOULD BE R OF X FOR THIS RECTANGLE,
WHICH WOULD JUST BE THE FUNCTION VALUE SINE SQUARED X.
ALSO NOTICE THAT THE WIDTH OF THIS RECTANGLE WOULD BE DELTA X,
WHICH IS THE REASON WHY WE'RE INTEGRATING WITH RESPECT TO X.
THIS DISK IS ACTUALLY A RIGHT CIRCULAR CYLINDER
WHICH WOULD HAVE THE VOLUME OF V = PI R SQUARED H.
WHICH WE CAN SEE IN OUR INTEGRAL FORMULA.
HERE'S THE PI R SQUARED IS R OF X SQUARED
AND DX OR DELTA X WOULD BE THE HEIGHT
OR H OF THE RIGHT CIRCULAR CYLINDER.
SO WE HAVE ALL THE INFORMATION WE NEED
IN ORDER TO SET UP OUR INTEGRAL.
THE VOLUME OF THE SOLID
IS EQUAL TO PI x THE INTEGRAL OF SINE SQUARED X SQUARED,
INTEGRATED WITH RESPECT TO X FROM 0 TO PI.
NOW LET'S EVALUATE THIS INTEGRAL ON THE NEXT SLIDE.
NOTICE HOW WE COULD WRITE THE INTEGRAND AS SINE TO THE 4th X
BUT BECAUSE WE HAVE AN EVEN POWER OF SINE,
WE'LL HAVE TO APPLY THE POWER REDUCING FORMULA
THEY'VE GIVEN HERE BELOW WHERE
SINE SQUARED U = THE QUANTITY OF 1 - COSINE 2U DIVIDED BY 2.
SO WE CAN WRITE THIS AS
PI x INTEGRAL OF 1 - COSINE 2X DIVIDED BY 2.
THIS IS ONE FACTOR OF SINE SQUARED
BUT BECAUSE WE HAVE SINE SQUARED X SQUARED
THIS WOULD BE SQUARED AS WELL
AND NOW WE'LL HAVE TO MULTIPLY THIS OUT.
NOTICE HOW OUR DENOMINATOR WOULD BE 2 SQUARED OR 4.
SO LET'S GO AHEAD AND FACTOR OUT THE 1/4
AND WRITE THIS AS PI/4 x INTEGRAL OF,
WE HAVE 1 - COSINE 2X TIMES ANOTHER FACTOR OF 1 - COSINE 2X.
NOW LET'S MULTIPLY THIS OUT. I'M GOING TO DO THIS BELOW.
SO WE'LL HAVE 4 PRODUCTS. ONE, TWO, THREE, AND FOUR.
SO WE HAVE ONE,
THEN WE HAVE -COSINE 2X, -COSINE 2X,
WE HAVE -2 COSINE 2X
AND THEN WE HAVE COSINE SQUARED 2X.
SO I CAN WRITE THIS AS
PI/4 x INTEGRAL OF 1 - 2 COSINE 2X + COSINE SQUARED 2X.
BUT NOW WE HAVE TO APPLY
THE POWER REDUCING FORMULA FOR COSINE SQUARED 2X,
WHICH IS ALSO GIVEN HERE BELOW.
NOTICE HOW WE ARE GOING TO DOUBLE THE ANGLE,
LET'S ALSO FACTOR OUT THE 1/2.
SO WE HAVE + 1/2 x THE QUANTITY 1 + COSINE 4X.
LET'S GO AHEAD AND DISTRIBUTE THE 1/2 HERE
AND COMBINE IN LIKE TERMS.
NOTICE HERE WE HAVE 1 AND THEN + 1/2,
SO THAT WOULD BE 3/2 - 2 COSINE 2X
AND THEN WE'LL HAVE + 1/2 COSINE 4X.
OF COURSE, ALL THESE ARE FROM 0 TO PI.
NOW WE'LL GO AHEAD AND FIND THE ANTI-DERIVATIVE.
WE WILL HAVE TO PERFORM U SUBSTITUTION
TO FIND THE ANTI-DERIVATIVE OF THE SECOND AND THIRD TERMS.
TO SAVE A LITTLE BIT OF TIME,
I'M NOT GOING TO BREAK THIS UP INTO THREE SEPARATE INTEGRALS
BUT NOTICE HOW FOR THE SECOND TERM IF WE LET U = 2X
THEN DIFFERENTIAL U WOULD BE 2DX
AND THEREFORE, WE DIVIDE BOTH SIDES BY 2.
NOTICE THAT DX IS EQUAL TO 1/2 TO U,
SO WE'LL HAVE AN EXTRA FACTOR OF 1/2
WHEN WE FIND THE ANTI-DERIVATIVE OF THIS TERM
AND THEN FOR THE THIRD TERM IF WE LET V = 4X, DV WOULD = 4DX.
IF WE DIVIDE BOTH SIDES BY 4 NOTICE HOW DX = 1/4 DV.
SO WE'LL HAVE AN EXTRA FACTOR OF 1/4
WHEN WE FIND THE ANTI-DERIVATIVE OF THIS THIRD TERM.
SO WE HAVE PI/4 x THE ANTI-DERIVATIVE OF 3/2
WITH RESPECT TO X WOULD BE 3/2X
AND THEN WE'LL HAVE - 2 TIMES THE ANTI-DERIVATIVE OF COSINE 2X
AGAIN WE'LL HAVE THIS EXTRA FACTOR OF 1/2
SO WE'LL HAVE 1/2 SINE 2X AND THEN + 1/2
x THE ANTI-DERIVATIVE OF COSINE 4X WOULD BE 1/4 SINE 4X.
AND THE LIMITS OF INTEGRATION WERE FROM 0 TO PI.
LET'S SIMPLIFY THIS ONE MORE TIME.
SO WE HAVE 3/2X THIS WOULD JUST BE - SINE 2X
AND HERE WE HAVE + 1/8 SINE 4X.
NOW WE'LL EVALUATE THIS AT PI AND 0
AND THEN FIND THE DIFFERENCE.
SO WHEN X = PI WE'D HAVE 3 PI/2.
HERE WE'D HAVE SINE 2 PI WHICH IS 0, SO - 0.
HERE WE WOULD HAVE SINE 4 PI WHICH IS 0, SO + 0,
AND THEN WHEN X IS 0 NOTICE ALL THREE TERMS WOULD BE 0.
SO WE'D HAVE 0 - 0 + 0.
SO THIS WORKS OUT TO PI/4 x 3 PI/2
SO THE EXACT LINE OF OUR SOLID
WOULD BE 3 PI SQUARED DIVIDED BY 8
PLUS ALSO WE GET A DECIMAL APPROXIMATION.
OUR NEW MEASURE IS 3 PI SQUARED AND THEN WE HAVE TO DIVIDE BY 8.
SO THE APPROXIMATE VOLUME WOULD BE 3.7011.
AGAIN, WE FOUND THE EXACT VOLUME AS 3 PI SQUARED DIVIDED BY 8
AND THE APPROXIMATE VOLUME OF 3.7011.
I HOPE YOU FOUND THIS HELPFUL.