[I got a mistake here, gimme a minute]
We'll muddle through without Examination Figure I. Are you supposed to be using Brainly for tests? I think I had an answer deleted once because the question came from a test.
Let's proceed. We'll center the parabola on the y axis, and use the x axis as the bottom of the piece we want. So the vertex is (0,4) and its zeros are (-6,0) and (6,0), giving a base of 12.
To have those zeros, our parabola must be an equation of the form
[tex]f(x)=a (x+6)(x-6) = a(x^2-36)[/tex]
We need f(0)=4 so a=4/(-36)=-1/9
[tex]f(x)=\frac 1 9 (x+6)(x-6) = \frac 1 9 (36 - x^2) = 4 - (\frac{x}{3})^2[/tex]
I'm assume without seeing the picture that the force is some function of the density and the area of the submerged part of the parabola section. We'll work out the area here; I think I'll need the picture to be sure what exactly they want you to do with the density.
Anyway, the area of the submerged bit is related to the integral of from -6 to 6. That integral counts the piece that's not submerged, which is another integral of f whose bounds we'll have to work out.
To find out the bounds where we're above water, we solve
[tex]f(x) = 2[/tex]
[tex]4 - (\frac{x}{3})^2 = 2[/tex]
[tex]-(\frac{x}{3})^2= -2[/tex]
[tex]\frac{x}{3} = \pm \sqrt 2[/tex]
[tex]x = \pm 3 \sqrt 2[/tex]
Those are our integration bounds. Our area is
[tex]A = \int_{-6}^{6} f(x) dx - \int_{-3 \sqrt 2}^{3 \sqrt 2} [/tex]
Let's work out the integral of f without the bounds first.
[tex]\int f(x) dx = \int (4 - \frac{x^2}{9} )dx[/tex]
[tex]\int f(x) dx = 4 x - x^3/27 = 4x - (x/3)^3[/tex]
We won't worry about the constant because we're doing definite integrals.
[tex]A =( 4x - (x/3)^3)|_{-6}^{6} - ( 4x - (x/3)^3)|_{-3 \sqrt 2}^{3 \sqrt 2} f(x) dx[/tex]
The integral is odd so we can just double the values at the right endpoints.
[tex]A =2 ( 4(6) - (6/3)^3) - ( 4 (3 \sqrt 2) -(3 \sqrt 2) /3)^3 )[/tex]
[tex]A =32 - 20 \sqrt 2[/tex]
Without the figure I'll leave the rest to you.
