- A TROUGH IS 6 FEET LONG AND ONE FOOT HIGH.
THE VERTICAL CROSS SECTION OF THE TROUGH PARALLEL TO THE END
IS SHAPED LIKE THE GRAPH OF Y = X SQUARED FROM X = -1 TO X = 1.
THE TROUGH IS FULL OF WATER.
WE WANT TO FIND THE AMOUNT OF WORK AND FOOT-POUNDS REQUIRED
TO EMPTY THE TROUGH BY PUMPING THE WATER OVER THE TROUGH.
IN THIS PROBLEM WE'LL USE 62 POUNDS PER CUBIC FOOT
AS THE WEIGHT OF WATER.
TO HELP US SET THIS PROBLEM UP
WE'LL FIRST DETERMINE
A SMALL INCREMENT OF WORK OR DELTA W SUB "I,"
WHICH WILL BE THE AMOUNT OF WORK TO PUMP OUT
1 LAYER OF WATER OVER THE TOP.
WHERE THE INCREMENT OF WORK IS EQUAL TO THE FORCE
x THE DISTANCE INCREMENT
AND THEN AS WE FIND THE SUM
OF MORE AND MORE INCREMENTS OF WORK
IT APPROACHES THE TOTAL WORK WHERE FINALLY IT LEADS
TO THE LIMIT AS N APPROACHES INFINITY OF THE SUM
OF THESE INCREMENTS OF WORK,
WHICH GIVES US OUR DIFF INTEGRAL.
NOTICE HERE WE'RE INTEGRATING WITH RESPECT TO X.
IN OUR EXAMPLE, WE'LL INTEGRATE WITH RESPECT TO Y.
BUT THE KEY IS DETERMINE THE AMOUNT OF WORK REQUIRED
TO PUMP THIS LAYER OF WATER OVER THE TOP.
WE'LL DETERMINE THIS INCREMENT OF WORK
BY DETERMINING THE FORCE OF THIS LAYER OF WATER
AND THEN MULTIPLY IT BY THE DISTANCE.
TO FIND THE FORCE,
WE'LL DETERMINE THE WEIGHT OF THIS WATER
AND SINCE WE'RE GIVEN THAT WE'RE GOING TO USE
62 POUNDS PER CUBIC FOOT AS THE WEIGHT
WE'LL BEGIN BY DETERMINING THE VOLUME OF THIS LAYER OF WATER
AND THEN MULTIPLY BY 62 POUNDS PER CUBIC FOOT
TO FIND THE WEIGHT WHICH WILL BE OUR TOTAL FORCE.
SO LOOKING AT OUR DIAGRAM HERE
NOTICE HOW THE THICKNESS OF THE LAYER WILL BE DELTA Y.
WE'LL CALL IT DELTA Y SUB "I."
THE DISTANCE THE WATER MUST BE PUMPED
WOULD BE THIS LENGTH HERE.
SO IF WE CALL THE LENGTH BELOW THIS Y SUB "I,"
THEN THE DISTANCE THE WATER MUST BE PUMPED
WOULD BE THIS DISTANCE HERE,
WHICH WILL BE THE TOTAL DISTANCE OF 1 - Y SUB "I."
NOW GOING BACK TO THE VOLUME OF THIS LAYER OF WATER,
WE KNOW THE THICKNESS IS DELTA Y SUB "I."
WE KNOW THIS LENGTH HERE IS THE SAME AS THIS LENGTH OF 6 FEET,
BUT WE DON'T KNOW THE WIDTH HERE.
SO WE'LL HAVE TO DETERMINE THIS WIDTH AS A FUNCTION OF Y.
TO DO THIS, LET'S LOOK AT THE GRAPH OF Y = X SQUARED
ON THE COORDINATE PLANE.
THE WIDTH WE'RE LOOKING FOR IS THIS WIDTH HERE
WHICH WE CAN SEE WOULD BE 2X.
BUT WE WANT TO EXPRESS THIS WIDTH AS A FUNCTION OF Y,
SO IF WE HAVE Y = X SQUARED OR X SQUARED = Y,
IF WE TAKE THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION
NOTICE HOW THIS GIVES US X = + OR - SQUARE ROOT Y.
WHICH MEANS THIS DISTANCE WOULD BE SQUARE ROOT Y
AND THIS DISTANCE WOULD ALSO BE SQUARE ROOT Y.
AND THEREFORE, THE TOTAL WIDTH WOULD BE 2 SQUARE ROOT Y.
SO IF THIS WIDTH IS 2 SQUARE ROOT Y
OR 2 SQUARE ROOT Y SUB "I"
WE HAVE ALL THE INFORMATION WE NEED
TO FIND THE VOLUME OF THIS LAYER OF WATER.
THIS IS A RIGHT RECTANGULAR PRISM OR A BOX
AND THEREFORE THE VOLUME IS EQUAL TO LENGTH x WIDTH x HEIGHT
WHERE THE LENGTH, WE ALREADY SAID IS 6 FEET.
THE WIDTH WE JUST FOUND AS 2 x SQUARE ROOT Y SUB "I,"
AND THE THICKNESS IS DELTA Y SUB "I."
AND THEREFORE THE FORCE OR THE WEIGHT
IS EQUAL TO 62 POUNDS PER CUBIC FOOT x THE VOLUME GIVEN HERE.
AND WE ALSO KNOW THAT THE DISTANCE
THAT THE WATER MUST TRAVEL IS 1 - Y SUB "I."
WHICH MEANS THE INCREMENT OF WORK REQUIRED
TO PUMP THIS LAYER OVER THE TOP, DELTA W SUB "I,"
WOULD BE EQUAL TO THE FORCE x THE DISTANCE WHICH WOULD BE 62,
WE'LL LEAVE OFF THE UNITS,
x 12 SQUARE ROOT Y SUB "I" x DELTA Y SUB "I"
x THE DISTANCE OF 1 - Y SUB "I."
WHICH MEANS THE TOTAL WORK WOULD BE EQUAL TO THE LIMIT
AS N APPROACHES INFINITY
OF THE SUM OF ALL THE INCREMENTS OF WORK
WHICH WOULD BE 62 x 12 SQUARE ROOT Y SUB "I"
DELTA Y SUB "I" x 1 - Y SUB "I."
AS FAR AS THE INTEGRAL,
WE HAVE THE TOTAL WORK IS EQUAL TO THE INTEGRAL OF AGAIN,
62 x 12 SQUARE ROOT Y.
NOW DELTA Y BECOMES DIFFERENTIAL Y,
AND WE HAVE THE QUANTITY 1 - Y,
AND WE'LL INTEGRATE FROM THE BOTTOM OF THE TROUGH TO THE TOP
OR FROM 0 TO 1.
NOTICE HERE 62 x 12 = 744
WHICH GIVES US THIS INTEGRAL HERE.
LET'S GO AHEAD AND FACTOR OUT THE 744.
LET'S ALSO WRITE SQUARE ROOT Y AS Y TO THE 1/2.
NOW LET'S DISTRIBUTE.
SO WE HAVE Y TO THE 1/2 - Y TO THE 3/2
AND THEN WE'LL FIND THE ANTI-DERIVATIVE.
THE ANTI-DERIVATIVE OF Y TO THE 1/2
WOULD BE Y TO THE 3/2 DIVIDED BY 3/2
OR 2/3 Y TO THE 3/2
AND - THE ANTI-DERIVATIVE OF Y TO THE 3/2
WOULD BE Y TO THE 5/2 DIVIDED BY 5/2 OR 2/5 Y TO THE 5/2
SO NOW WE HAVE 744 x--
WHEN Y IS 1 WE JUST HAVE 2/3 - 2/5
AND WHEN Y IS 0 THESE ARE BOTH 0.
SO NOW LET'S GO TO THE CALCULATOR.
WE HAVE (744 x 2/3) - 2/5, WHICH GIVES US 198.4
AND THIS WOULD BE FOOT-POUNDS.
THIS IS THE AMOUNT OF WORK REQUIRED
TO PUMP THE WATER OUT OF THE TROUGH.
I HOPE YOU FOUND THIS HELPFUL.