Asked by yasmin
can some one help me with this question
The inner surface of a bowl is of the shape formed by rotating completely about the y- axis the
area bounded by the curve 𝑦 = 𝑥^2 − 4 , the x – axis, the y – axis and the line 𝑦 = 3 .
(i) Find the volume of the bowl
(ii) Find the formula that represents the volume of water in the bowl when the depth of water is d (< 3)
If water is poured in at a rate of 5 cubic units per second
(iii) Find the rate at which the depth is increasing when 𝑑 = 1.
The inner surface of a bowl is of the shape formed by rotating completely about the y- axis the
area bounded by the curve 𝑦 = 𝑥^2 − 4 , the x – axis, the y – axis and the line 𝑦 = 3 .
(i) Find the volume of the bowl
(ii) Find the formula that represents the volume of water in the bowl when the depth of water is d (< 3)
If water is poured in at a rate of 5 cubic units per second
(iii) Find the rate at which the depth is increasing when 𝑑 = 1.
Answers
Answered by
oobleck
(i) using discs of thickness dy,
v = ∫[0,3] πr^2 dy
where r = x = √(y+4)
v = ∫[0,3] π(y+4) dy = 33π/2
(ii) v = ∫[0,d] π(y+4) dy = π(d^2/2 + 4d)
(iii) dv/dt = π(d+4) dd/dt
so plug in your numbers
check on (i) using shells of thickness dx
v1 = 12π
v2 = ∫[2,√7] 2πx(3-(x^2-4) dx = 9π/2
v = 12π + 9π/2 = 33π/2
v = ∫[0,3] πr^2 dy
where r = x = √(y+4)
v = ∫[0,3] π(y+4) dy = 33π/2
(ii) v = ∫[0,d] π(y+4) dy = π(d^2/2 + 4d)
(iii) dv/dt = π(d+4) dd/dt
so plug in your numbers
check on (i) using shells of thickness dx
v1 = 12π
v2 = ∫[2,√7] 2πx(3-(x^2-4) dx = 9π/2
v = 12π + 9π/2 = 33π/2
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