Answer:
[tex]pg\int\limits^7_7(10-y)2\sqrt{7^2-y^2} \, dy[/tex]
*The last bound is negative 7
Then it equals 15085928 or 1.5E7
Step-by-step explanation:
[tex]x^2+y^2=7^2[/tex]
area = 2xdy
depth = (7+3)-y -> (10-y)
total force = [tex]pg\int\limits^7_7 {(10-y)2x} \, dy[/tex]
Substitute 2x from the first equation as x=[tex]\sqrt{7^2-y^2}[/tex]
total force = [tex]pg\int\limits^7_7 {(10-y)2\sqrt{49-y^2} } \, dy[/tex]
=pg1539.38
=(1,000)(9.8)(1539.38)=1.5E7 N
*Lower bound is -7, can't get the program to allow me to put it as a negative
The hydrostatic force against one side of the plate is an integral 1.5E7 N.
The general equation of a circle is [tex](x-h)^{2} +(y-k)^{2} = r^{2}[/tex]. The coordinate is (0, 0) at the origin, so the equation will become [tex](x)^{2} +(y)^{2} = r^{2}[/tex].
[tex](x)^{2} +(y)^{2} = 7^{2}[/tex]
area of circular plate = 2xdy
depth = (7+3)-y = (10-y)
The total force = [tex]pg\int\limits^7_7[ {10-y} ]2xdy[/tex]
Substitute 2x from the first equation as x = [tex]\sqrt{7^{2} -y^{2} }[/tex][tex]pg\int\limits^7_7[ {10-y} ]\sqrt{49 -y^{2} }dy[/tex]
= pg 1539.38
= (1,000)(9.8)(1539.38)
= 1.5E7 N
Thus, The hydrostatic force against one side of the plate is an integral 1.5E7 N.
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