Suppose a tank is created by revolving the region R bounded by graphs of x=g(y),

Question

Suppose a tank is created by revolving the region R bounded by graphs of x=g(y), x=0, y=0, and y=H about the y-axis, where g(y)>0 on [0,H]. Show that if the tank is filled to depth h, then the volume of the water in the tank is V(h)= (pieint_{0}^{2}g(y)^2 dy.) (int_{?}^{?}int_{?}^{?})(int_{?}^{?}int_{?}^{?}) Find the volume function V=f(h), for the tank formed when the region R bounded by the graphs y=x+1, x=0, y=0, and y=4 is revolved about the y axis. Find the volume function V=f(h), for the tank formed when the region R bounded by the graphs of x=y^2+1, x=0, y=0, and y=3 is revolved about the y axis.

Explanation / Answer

1) Area = pi*x^2 dy = pi* (g(y))^2 dy

Volume = [int 0 to h] pi* (g(y))^2 dy

= pi * [int 0 to h] (g(y))^2 dy

2) x = y - 1

Volume = pi * [int 0 to h] (g(y))^2 dy

= pi*[int 0 to h] (y - 1)^2 dy

= pi*[int 0 to h] (y^2 - 2y + 1) dy

= pi* (y^3 / 3 - y^2 + y) [0 to h]

= pi* (h^3 - 3h^2 + h) / 3

3) x = y^2 + 1

Volume = pi * [int 0 to h] (g(y))^2 dy

= pi*[int 0 to h] (y^2 + 1)^2 dy

= pi*[int 0 to h] (y^4 + 2y^2 + 1) dy

= pi* (y^5 / 5 + 2y^3 / 3 + y) [0 to h]

= pi* (h^5 / 5 + 2h^3 / 3 + h)

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