Calculus 2 - volume of water in sphere

A hemispherical bowl of radius r contains water to a depth h. Give a formula that you can use to measure the volume of the water in the bowl.

I'm thinking that each cross section will be pi r^2... And the integral will be from 0 to h. Am I missing something?

take a cross-section with radius x . this cross-section will have very very small height(Δy) and will resemble a slice (cylinder)

to find the volume of each cylinder use πx²(Δy)

to find the total volume of the water in the bowl just integrate all them cylinders from a height of 0 to a height of h.

V = ∫πx² dy
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NOTE: the radius of the bowl, r, will never vary unless the bowl goes under deformation. so r is constant but the height of the water will vary

but you cannot integrate directly a changing x with respect to a changing y. determine the relationship between x and y and eliminate one of them. in this case try to eliminate r .

relationship between x and y
```````````````` `````````````````` ```imagine the bow sitting on the table. let the surface of the table to be x-axis and the height of the ball to be the h-axis; let the base of the bowl be the origin(0,0).
(x - 0)² + (y - r)² = r²
x² + y² - 2ry + r² = r²
x² = 2ry - y²
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volume:
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V = ∫πx² dy from 0 to h
V = ∫π( 2ry - y²) dy
V = π ∫( 2ry - y²) dy
V = π (ry² - ⅓ y³)
V = π (rh² - ⅓ h³)
V = πh² (r - ⅓ h)
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