- WE WANT TO FIND THE VOLUME 
OF THE SOLID
OBTAINED BY ROTATING THE REGION 
BOUNDED BY Y = 3X SQUARED,
X = 1, X = 3, AND Y = 0 
ABOUT THE X AXIS.
AND WE'LL DO THIS 
USING THE DISK METHOD.
SO FIRST HERE'S THE GRAPH 
OF Y = 3X SQUARED.
HERE'S THE GRAPH OF X = 1.
HERE'S THE GRAPH OF X = 3, 
AND Y = 0 WOULD BE THE X AXIS.
SO THE BOUNDED REGION 
WOULD BE THIS REGION HERE.
IF WE ROTATE THIS REGION ABOUT 
THE X AXIS,
NOTICE HOW IT WOULD 
FORM THIS SOLID HERE,
AND OUR GOAL IS TO FIND 
THE VOLUME OF THIS SOLID.
WE'LL DO THIS USING THE FORMULA 
V = PI x THE INTEGRAL OF F OF X
SQUARED FROM "A" TO B.
BEFORE WE APPLY THIS, THOUGH,
LET'S LOOK AT WHERE 
THIS COMES FROM.
IF WE WERE TO SKETCH 
A SMALL THIN RECTANGLE
OF THIS BOUNDED REGION 
LIKE THIS,
WHERE THIS WOULD BE DELTA X
AND THE HEIGHT WOULD BE F OF X 
OF THE FUNCTION VALUE.
IF YOU ROTATE THIS 
ABOUT THE X AXIS,
NOTICE HOW THIS WOULD BE A VERY 
THIN RIGHT CIRCULAR CYLINDER
THAT WOULD LOOK SOMETHING 
LIKE THIS.
WE KNOW THE VOLUME OF A RIGHT 
CIRCULAR CYLINDER
IS EQUAL TO PI R SQUARED H,
WHERE IN THIS CASE THIS DISTANCE 
HERE WOULD BE R, THE RADIUS,
AND THIS DISTANCE HERE 
WOULD BE THE HEIGHT.
BUT NOTICE HOW WE CAN SEE THIS 
FORMULA IN THE INTEGRAL
WE'RE USING TO FIND THE VOLUME.
HERE'S THE PI.
R SQUARED, OR THE RADIUS 
SQUARED,
IS ACTUALLY GOING TO BE THE SAME 
AS THE FUNCTION VALUE SQUARED,
AND THEN THE HEIGHT OF THE RIGHT 
CIRCULAR CYLINDER H
IS ACTUALLY DX, OR DELTA X.
SO TO VISUALIZE THIS, IF WE 
ROTATE A REGION ABOUT THE X AXIS
IT MIGHT LOOK 
SOMETHING LIKE THIS.
TO APPROXIMATE THIS VOLUME
WE COULD USE
RIGHT CIRCULAR CYLINDERS, 
AS WE SEE HERE.
BUT NOTICE AS THE NUMBER 
OF RIGHT CIRCULAR CYLINDERS
INCREASES 
AND APPROACHES INFINITY,
THE VOLUME OF THE DISK, 
OR RIGHT CIRCULAR CYLINDERS,
WOULD APPROACH 
THE VOLUME OF THE SOLID.
SO AS THE NUMBER OF DISKS 
APPROACH INFINITY,
THIS FORMS OUR INTEGRAL THAT 
WE'RE USING TO FIND THE VOLUME.
SO GOING BACK TO OUR EXAMPLE, 
LET'S GO AHEAD AND SET THIS UP.
WE'RE GOING TO HAVE THE VOLUME
= PI x THE INTEGRAL OF F OF X 
SQUARED,
WHICH WOULD BE 3X TO THE SECOND 
SQUARED FROM 1 TO 3.
SO THIS IS GOING TO BE PI x THE 
INTEGRAL OF,
THIS WOULD BE 9X TO THE FOURTH 
FROM 1 TO 3.
LET'S GO AHEAD AND EVALUATE 
THIS ON THE NEXT SLIDE.
THE ANTIDERIVATIVE OF 9X 
TO THE FOURTH
WOULD BE 9 x X TO THE FIFTH 
DIVIDED BY 5.
LET'S GO AHEAD 
AND FACTOR OUT THE 9/5.
THIS WOULD GIVE US 9PI 
DIVIDED BY 5.
AND NOW WE'LL EVALUATE X TO THE 
FIFTH AT 3 AND THEN 1,
AND THEN FIND THE DIFFERENCE.
SO WHEN X IS 3, 
WE HAVE 3 TO THE FIFTH,
AND WHEN X IS 1, 
WE HAVE 1 TO THE FIFTH.
WELL, 3 TO THE FIFTH IS 243.
1 TO THE FIFTH IS 1.
SO THIS GIVES US 242.
9PI/5 x 242 = 2,178PI 
DIVIDED BY 5,
WHICH WOULD BE 
THE VOLUME OF THE SOLID,
SO THIS WOULD BE CUBIC UNITS.
LET'S ALSO GET A DECIMAL 
APPROXIMATION THOUGH.
THIS WOULD BE APPROXIMATELY 
1,368.478 CUBIC UNITS.
FOR OUR HOMEWORK, THOUGH,
WE SHOULD ENTER THE EXACT VALUE 
UNLESS WE'RE TOLD TO ROUND.
SO, AGAIN, THIS IS THE VOLUME 
OF THIS SOLID HERE.
I HOPE YOU HAVE FOUND 
THIS HELPFUL.
