No.

1st, with your (3.14)(25) for area of circular disc, you are are using only the top of the hemisphere. Your dA should be any random circular disc below that topmost circle. Your dA should be (3.14)(x^2), where x is the radius of that random disc.

2nd, your integration from 0 to 10 for dy means you were thinking the hemisphere were a sphere. The boundaries of dy in the hemisphere is from 0 to 5 only, if the origin (0,0) is at the bottom of the bowl. It is from (-5) to 0, if (0,0) is at the center of the topmost disc.

Draw the figure.

It is a semi-circle, such that the topmost, a diameter, is horizontal and is 2(5) = 10ft wide. Its midpoint is (0,0), the center of the circle.

Draw a horizontal chord, the surface of the water in the bowl, below that topmost line. Draw a radius to one of the ends of that chord.

A right triangle is formed, with these:

---vertical leg = -y

---horizontal leg = x

---hypotenuse = r = 5ft

By Pythagorean theorem,

x^2 +y^2 =25

x^2 = 25 -y^2

dA = pi(x^2) = pi(25 -y^2) -----------------sq.ft.

dV = (dA)(dy) = pi(25 -y^2)dy --------------cu.ft.

dWeight = dV*62.5 = (62.5pi)(25 -y^2)dy ----lbs.

To get that dWeight out of the bowl, it must be raised up the (-y) vertical distance from the top, so,

dWork = (-y)(dWeight) = (-62.5pi)(25 -y^2)(y dy)

Integration for dy is from y = -5 to y=0, so,

Work = (-62.5pi)INT.(-5 --> 0)[25 -y^2](y dy)

Work = (-62.5pi)INT.(-5 --> 0)[25 -y^2](y dy)(-2/(-2))

Work = (-62.5pi/(-2))INT.(-5 --> 0)[25 -y^2](-2y dy)

Work = (31.25pi)[(1/2)(25 -y^2)^2]|(-5 --> 0)

Work = (15.625pi)[(25 -(0)^2)^2 -(25 -(-5)^2)^2]

Work = (15.625pi)[625]

Work = 9765.625pi = 30,680 ft-lbs --------answer.