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— A CIRCULAR SWIMMING POOL HAS A DIAMETER OF 14 METERS. THE SIDES ARE 4 METERS HIGH AND THE DEPTH OF THE WATER IS 3 METERS. HOW MUCH WORK IN JOULES IS REQUIRED TO PUMP All the WATER OVER THE SIDE? THE ACCELERATION DUE TO GRAVITY IS 9.8 METERS PER SECOND SQUARED AND THE DENSITY OF WATER IS 1,000 KILOGRAMS PER METERS CUBED. SO FOR A QUICK REVIEW, IN GENERAL, WORK IS EQUAL TO FORCE TIMES DISTANCE BUT IF AN OBJECT IS MOVING ALONG A STRAIGHT LINE BY CONTINUOUSLY VARYING FORCE F OF X THEN THE WORK DONE BY THE FORCE AS THE OBJECT IS MOVED FROM X EQUALS quot;Pequot; TO X EQUALS B IS GIVEN BY THE DIFF INTEGRAL OF F OF X, INTEGRATED WITH RESPECT TO X FROM quot;Pequot; TO B. TO UNDERSTAND WHY THIS WORKS, IF WE CONSIDER 1 INCREMENT OF WORK OR IN OUR CASE, THE AMOUNT OF WORK REQUIRED TO PUMP 1 DISK OF WATER OVER THE TOP, IT WOULD BE EQUAL TO THE FORCE TIMES THE DISTANCE INCREMENT. AS WE FIND MORE AND MORE INCREMENTS OF WORK AND THEN SOME OF THOSE INCREMENTS WOULD APPROACH TO YOUR TOTAL WORK AND THEREFORE THE WORK IS EQUAL TO THE LIMIT AS N APPROACHES INFINITY OF THE SUM OF DELTA W SUB quot;Pequot; FROM quot;Pequot; = 1 TO N. AND THIS LIMIT GIVES US OUR DIFF INTEGRAL. SO TO HELP US SET UP OUR INTEGRAL, WE'LL FOCUS ON DETERMINING 1 INCREMENT OF WORK WITH THE AMOUNT OF WORK REQUIRED TO PUMP 1 DISK OVER THE TOP AND THIS WILL HELP US SET UP OUR DIFF INTEGRAL. SO GOING BACK TO OUR EXAMPLE, LET'S LABEL THE GIVEN INFORMATION. THIS CIRCULAR SWIMMING POOL HAS A DIAMETER OF 14 METERS AND THEREFORE THE RADIUS WOULD BE EQUAL TO HALF THIS OR 7 METERS. NEXT, THE SIDES ARE 4 METERS HIGH WHICH IS LABELED HERE. THE DEPTH OF THE WATER IS 3 METERS LABELED HERE. IF WE CONSIDER THIS DISK OF WATER HERE, NOTICE HOW THE HEIGHT OR THE THICKNESS WOULD BE DELTA X OR MORE SPECIFICALLY DELTA X SUB quot;Pequot; AND IF YOU LET X BE THE DISTANCE FROM THE BOTTOM THEN WE COULD LABEL THIS LENGTH HERE X SUB quot;Pequot; AND THEREFORE NOTICE HOW THIS DISK MUST TRAVEL 4 — X SUB quot;Pequot; METERS OVER THE TOP. NOW AGAIN, OUR GOAL HERE IS TO FIND THE INCREMENT OF WORK REQUIRED TO PUMP THIS DISK OF WATER OVER THE TOP WHICH WE'LL CALL DELTA W SUB quot;Pequot; WHICH IS EQUAL TO THE FORCE x THE DISTANCE. BUT BEFORE WE FIND THE FORCE WE HAVE TO FIND THE VOLUME OF THIS DISK AND SINCE THIS DISK IS A RIGHT CIRCULAR CYLINDER THE VOLUME = PI R SQUARED H, WHERE R IS 7 AND THE HEIGHT, H = DELTA X SUB quot;I.quot; THEREFORE THE VOLUME = 49 PI x DELTA X OF I. AND NOW WE KNOW FORCE = MASS x ACCELERATION BUT WE DON'T KNOW THE MASS OF THIS DISK OF WATER. HOWEVER, SINCE THE MASS = THE DENSITY x THE VOLUME AND WE KNOW BOTH OF THESE WE CAN NOW FIND THE MASS. THE DENSITY OF THE WATER IS 1,000 KILOGRAMS PER METER SQUARED. THE VOLUME WE JUST FOUND AND ACCELERATION DUE TO GRAVITY, quot;A, IS 9.8 METERS PER SECOND SQUARED. SO THIS PRODUCT GIVES US THE FORCE OF THIS DISK OF WATER AND NOW THE DISK OF WATER MUST TRAVEL AGAIN, 4 — X SUB quot;Pequot; METERS AND NOW WE CAN FIND THE INCREMENT OF WORK REQUIRED TO PUMP THIS ONE...
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