Need help with this calc 2 problem?

Question

A trough is 3 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{2} from x=-1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.

Answer 1

y = x^2

integral of y [1/3x^3]between 1 and -1

=  1/3 - (-1/3)

= 2/3

volume of trough = 3 x 1 x 2/3

=2

force required =62 x 2

= 124N

W=Fx

then i get stuck ;)

  

Answer 2

Work = Force * distance.  In this case, Force = the weight of the water that is moved, and distace = 1 - (actual hight of a particular small volume of water) since we are moving the water from its actual height, up to y=1.

The incremental volume, dV = l*w*h

dV = (3 ft)(x)(dy)

But we can't integrate both x & dy in 1-variable calculus, so we need to convert one of those variables into the other.  It makes sense to turn the x into a y, since we are moving our incremental volume from y=0 up to y=1.  We know y=x^2, so x = sqrt(y)

dV = 2*(3 ft)(sqrt(y))(dy) (I used 2* due to symmetry, since we are needing the width from -x to +x, which is just a 2*x distance, or 2*sqrt(y))

That gives us the volume of water, but we don't care about "volume".  We care about "weight".  We need to convert volume into weight, which is just weight = density * volume (force actually is equal to mass * acceleration, which in the US measurement system is just equal to weight)

dF = (62 pounds / ft^3)(2)(3 ft)(sqrt(y))(dy)

Now the distance the water is moved is (1-y)

so the incremental work = incremental force * distance

dW = (1-y)(62 pounds / ft^3)(2)(3 ft)(sqrt(y))(dy)

W = 372  Ã¢ÂˆÂ« (from y=0 to 1) (1-y)√(y) dy

W = 372  Ã¢ÂˆÂ« (from y=0 to 1) y^(1/2)-y^(3/2) dy

W = 372 [(2)y^(3/2)-(2/5)y^(5/2)] (from y=0 to 1)

W = 372 [2-2/5]

W = 372 [ 1.6 ] = 595.2