# Thread: work by pumping

1. ## work by pumping

A trough is 2 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^6 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.

ok the part thats screwing me up is The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^6 from x=–1 to x=1 im not sure what to do with that.

2. Originally Posted by dat1611
A trough is 2 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^6 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.

ok the part thats screwing me up is The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^6 from x=–1 to x=1 im not sure what to do with that.
volume of a horizontal cross-section is $2 \cdot 2x \cdot dy$

since $y = x^6$ , $x = y^{\frac{1}{6}}$

volume of a horizontal cross-section is $4y^{\frac{1}{6}} \cdot dy$

weight of a horizontal cross-section is $62 \cdot 4y^{\frac{1}{6}} \cdot dy$

distance each horizontal cross-section needs to be lifted is $(1-y)$

work = $\int_0^1 62 \cdot 4y^{\frac{1}{6}}(1-y) \, dy
$