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[west reef]

you can use integrals for this if you know how

you have to find the equation for the half circle with radius 12.2 and intersections with the x axis at (5.2,0) and (-5.2,0)

and the parabola with a vertex on the y axis at y=4.89, and the same x intercepts as the circle

and then integrate with respect to x from -5.2 to 5.2

[compact girder]

i know how to take an integral but im having troubles with the equation

i know half circle is

piR^2/2

[west reef]

the equation for a circle is x^2 + y^2 = r^2

so in this case x^2 + y^2 = (12.2)^2

from there you have to rearrange the equation so that it's y = f(x) and then translate the graph down such that it intersects the x axis at the given points

[compact girder]

so i have to bring it down by 12.2

[west reef]

to find the translation down needed for the correct x intercepts plug in x = 5.2 and subtract the resulting y output

[compact girder]

into sqrt of 12.2^2-(5.2)^2

[west reef]

yes

[compact girder]

i got 121.8

[west reef]

I think you just didn't square root it

you got the square of the correct value

[compact girder]

thats right

so i drop it by 11.03630

[west reef]

yes

[compact girder]

i see now i gotta take the integral between them

?

[west reef]

yeah so the integral of the top curve minus the bottom curve

from -5.2 to 5.2

[compact girder]

one sec lemme work that out

so the distance from the x-axis to the top of the circle is the 11.036?

[compact girder]

lil confused

so the distance from the x-axis from the top of the circle

would be 11.036

or?

[west reef]

no

[west reef]

the distance to the top of the circle from the x axis in the original form (in the message I responded to) is the radius of the circle 12.2 but in the original figure we see that the parabola and circle intersect twice 10.4 units apart from one another so to make the calculation easier we're just making that the distance between where the 2 graphs intersect the x axis

so when we subtracted the f(5.2) value for the translation down the goal was to bring the graph down so that it intersects with the x axis AT -5.2 and 5.2

and then the parabola should intercept at those points as well

[compact girder]

does it change anything if the dotted line is the y axis

i just have no clue how its 1.23 on this one

[west reef]

no that's actually what we've been doing so far

[compact girder]

so how did they get the 1.23 out of this?

is our like 5.2?

@west reef

[west reef]

I see that your problem has changed a bit so I'll show you this

[compact girder]

wheres the distance from top of circle on this?

[west reef]

the integral for this is:

[west reef]

not 100% sure what you mean

[compact girder]

so that finds me the area?

if i take the integral from -5.2 to 5.2

[west reef]

yeah

[compact girder]

where you get 5.2?

and 27.04

[compact girder]

??

[compact girder]

and 27.04

[compact girder]

?

[west reef]

that was the original value of f(0) for (x+5.2)(x-5.2) and I used it to solve for the correct coefficient for the parabola

[compact girder]

so what would be the equation for just the cirlce??

[compact girder]

??

.solved

[gray sky]

Solved

Post marked as solved by @compact girder.

Use .unsolved if this was a mistake.

[west reef]

lemme walk you through how to find each of the graphs.

for the circle you start with x^2 + y^2 = r^2
and then you convert it into a function of x like y = sqrt(r^2 - x^2)
then you figure out how much to translate it down by so that it intersects the x axis at the correct points. in the example we did earlier the total width from one bottom point to another was 10.4 ft which meant that each one of those points was 5.2 feet away from the y axis. so to get the amount that you subtract it by to translate it down to the correct level you have to plug in the x intercept value you want into the original equation and subtract by that much because you're moving the graph down by that exact amount so that it's zero at those points.

for the parabola you take the 2 x intercepts and you make a factorized polynomial in the previous case it was -(x+5.2)(x-5.2) because that way when x is 5.2 or -5.2 the value of the equation is 0, but that wasn't the actual equation for the graph because we needed the vertex of the parabola to be at y = 4.89 (at x = 0) so in order to get the value that we have to multiply the polynomial by we first set it to some unknown value a. we then have an equation that looks like this -a(0+5.2)(0-5.2) = 4.89 and you can use that to get "a" which was 4.89/27.04.

[compact girder]

ty