Answer:
[tex]62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb[/tex]
Explanation:
For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).
With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:
[tex]\int\limits^a_b[/tex]γhxdy
we know that γ=62.5 lb/[tex]ft^{3}[/tex]
from the drawing, we can determine the height (or depth under the water) of each differential area is given by:
h=8-y
x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}[/tex]
when substituting the x and y-values given on the graph, we get that the slope is:
[tex]m=\frac{0-6}{7-0}=-\frac{6}{7}[/tex]
once we got this slope, we can substitute it in the point-slope form of the equation:
[tex]y_{2}-y_{1}=m(x_{2}-x_{1})[/tex]
which yields:
[tex] y-6=-\frac{6}{7}(x-0)[/tex]
which simplifies to:
[tex] y-6=-\frac{6}{7}x[/tex]
we can now solve this equation for x, so we get that:
[tex]x=-\frac{7}{6}y+7[/tex]
with this last equation, we can substitute everything into our integral, so it will now look like this:
[tex]\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy[/tex]
Now that it's all written in terms of y we can now simplify it, so we get:
[tex]62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy [/tex]
we can now proceed and evaluate it.
When using the power rule on each of the terms, we get the integral to be:
[tex]62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}[/tex]
By using the fundamental theorem of calculus we get:
[tex]62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))][/tex]
When solving we get:
[tex]62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb[/tex]
