Question

Consider a tank which has the shape of an upside down pyramid with a rectangular base such that the length is twice the width W, and with height H, as shown above. The tank is filled to a depth of h ft from the bottom with a liquid of density 7070lb/ft33(assume that 0<h<H0<h<H).

Set up, but DO NOT SOLVE, an integral to find the work needed to empty the tank. (Your answer may include the constants WW, HH, and hh.)

Answer 1

please check density if it is 7070 or just 70, if later the case, answar can be obtained by just replacing 7070 with 70. thank you.

22 Hh xI. ce solution giver, density of liquid = 7070 18 1ff3 Let a strip of width dat at & from bottom Volume of strup, du= w. 2. ex In DOAB, w z w - w using in o du= 22 W² dx a ② force to pump, de ligued out of tork, - F=density & du = 7010 .22 W² da н u work done, dW=Fd, dis distance from top d= H-x 2. dw= 7070.22 W2 (4-2) da Integrating from x=0 to x = h we will get work done to empty the took, H H W = 7070 w² 20 (H-x) dx Ane H