- #1

nameVoid

- 241

- 0

placing one end of the tank at the origin and using rectangular cross sections. I have attempted to first calculate the bottom half for negative y values and then add it to the top

[tex]

C=9.8*10^3

[/tex]

[tex]

C\int_{-\frac{3}{2}}^{0}12\sqrt{\frac{9}{4}-y^2}(\frac{5}{2}+y)dy +C\int_{0}^{\frac{3}{2}}12\sqrt{\frac{9}{4}-y^2}(\frac{5}{2}-y)dy

[/tex]

also similar case here, A tank with semicircle base radius 2ft and height 8ft lies horizontl and has spout 1ft

placing the center of the circle at the orgin

[tex]

C=62.5

[/tex]

[tex]

C\int_{-2}^{0}16\sqrt{4-y^2}(1+y)dy

[/tex]

it looks likes the text is taking the limits of integration here from 0 to 2 I dont understand why