8. -/0.81 points details my notes ask your teacher practice another a circular plate with radius 6 m is…

8. -/0.81 points details my notes ask your teacher practice another a circular plate with radius 6 m is submerged vertically in water as shown. express the hydrostatic force against one side of the plate as an integral and evaluate it. (round your answer to the nearest whole number. use 9.8 m/s² for the acceleration due to gravity. recall that the mass density of water is 1000 kg/m³.) 6 5 m 9. 0.81/0.81 points details my notes previous answers ask your teacher a trapezoidal plate with height h and bases 9 and 2a is submerged vertically in water as shown. express the hydrostatic force against one side of the plate as an integral and evaluate it. (use delta for 6. assume a = 1, h = 1, and delta = 1000 kg/m³.)

8. -/0.81 points details my notes ask your teacher practice another a circular plate with radius 6 m is submerged vertically in water as shown. express the hydrostatic force against one side of the plate as an integral and evaluate it. (round your answer to the nearest whole number. use 9.8 m/s² for the acceleration due to gravity. recall that the mass density of water is 1000 kg/m³.) 6 5 m 9. 0.81/0.81 points details my notes previous answers ask your teacher a trapezoidal plate with height h and bases 9 and 2a is submerged vertically in water as shown. express the hydrostatic force against one side of the plate as an integral and evaluate it. (use delta for 6. assume a = 1, h = 1, and delta = 1000 kg/m³.)

Answer

Explanation:

Step1: Find the equation of the circle

The equation of the circle with radius (r = 6) is (x^{2}+y^{2}=36), so (x=\sqrt{36 - y^{2}}).

Step2: Determine the depth and width

The depth of a horizontal strip at height (y) is (d=(5 + y)) (since the top - most point of the circle is (y=- 5) from the water - surface reference). The width of the horizontal strip is (w = 2\sqrt{36 - y^{2}}).

Step3: Set up the hydro - static force integral

The formula for hydro - static force (F=\int_{a}^{b}\rho g d w dy), where (\rho = 1000\space kg/m^{3}), (g = 9.8\space m/s^{2}). Substituting the values of (d), (w), (\rho), and (g) into the formula, we get (F=\int_{-5}^{1}1000\times9.8\times(5 + y)\times2\sqrt{36 - y^{2}}dy).

Step4: Split the integral

(F = 19600\int_{-5}^{1}(5\sqrt{36 - y^{2}}+y\sqrt{36 - y^{2}})dy=19600\left(5\int_{-5}^{1}\sqrt{36 - y^{2}}dy+\int_{-5}^{1}y\sqrt{36 - y^{2}}dy\right)) For (\int y\sqrt{36 - y^{2}}dy), use the substitution (u = 36 - y^{2}), (du=-2ydy). (\int y\sqrt{36 - y^{2}}dy=-\frac{1}{2}\int\sqrt{u}du=-\frac{1}{3}u^{\frac{3}{2}}+C=-\frac{1}{3}(36 - y^{2})^{\frac{3}{2}}+C) (\int_{-5}^{1}y\sqrt{36 - y^{2}}dy=\left[-\frac{1}{3}(36 - y^{2})^{\frac{3}{2}}\right]{-5}^{1}=-\frac{1}{3}(36 - 1)^{\frac{3}{2}}+\frac{1}{3}(36 - 25)^{\frac{3}{2}}=-\frac{1}{3}(35^{\frac{3}{2}}-11^{\frac{3}{2}})) For (\int\sqrt{36 - y^{2}}dy), use the formula (\int\sqrt{a^{2}-y^{2}}dy=\frac{y}{2}\sqrt{a^{2}-y^{2}}+\frac{a^{2}}{2}\sin^{-1}(\frac{y}{a})+C) (here (a = 6)) (5\int{-5}^{1}\sqrt{36 - y^{2}}dy=5\left[\frac{y}{2}\sqrt{36 - y^{2}}+18\sin^{-1}(\frac{y}{6})\right]{-5}^{1}) (=5\left(\frac{1}{2}\sqrt{35}+18\sin^{-1}(\frac{1}{6})+\frac{5}{2}\sqrt{11}-18\sin^{-1}(-\frac{5}{6})\right)) After calculating (19600\left(5\int{-5}^{1}\sqrt{36 - y^{2}}dy+\int_{-5}^{1}y\sqrt{36 - y^{2}}dy\right)) using a calculator: (5\int_{-5}^{1}\sqrt{36 - y^{2}}dy\approx5\left(\frac{\sqrt{35}}{2}+18\times0.167+\frac{5\sqrt{11}}{2}+18\times0.985\right)) (\approx5\left(2.96 + 3.006+8.39+17.73\right)\approx5\times32.086 = 160.43) (\int_{-5}^{1}y\sqrt{36 - y^{2}}dy\approx-\frac{1}{3}(35^{\frac{3}{2}}-11^{\frac{3}{2}})\approx-\frac{1}{3}(207.06 - 36.48)\approx - 56.86) (F=19600\times(160.43-56.86)=19600\times103.57) (F = 2030072\approx2030072\space N)

Answer:

(\int_{-5}^{1}19600(5 + y)\sqrt{36 - y^{2}}dy), (2030072\space N)