Can someone help me figure out the area for this please?
>>8121346
Yes
>>8121346
>ft
no
>>8121346
double integral
dy from x^2 to sqrt(7)
dx from 0 to 4.5
>>8121346
Is "ft" the unit or is something else. Because if it's the unit (feet) the area can't be calculated
>>8121346
so hows that curve defined? there exists an infinite number of circles going through two points
>>8121364
It's obviously not a circle..
>>8121346
So the vertical distance is 20 ft?
>>8121368
if it's so obvious, what is it then?
>>8121357
fuck
ignore this
so i noticed the what seems like x^2 proportion on the lower right curve then formed a function from it using another variable
the function i got was [math] \frac{28x^2}{81} = 7 [/math]
then i integrated that function on the bounds and got 10.5
then i took the area of the rectangle and subtracted 10.5 from it
4.5*7 - 10.5 = 21
so it has an area of 21 feet
>>8121389
= 0 rather
>>8121346
Unless you give us more information such as the function which defines the curve, the best we can do is bound the area from below using convexity.
>>8121381
False dichotomy mate :)
It's obvious what it isn't, but not what it is.
>>8121409
I like you
>>8121389
Feet is a unit of area now?
>>8121346
No, because we don't know what the curve is between its two points. What we can say though is that the figure is bigger than the biggest right triangle contained in it (15.75), and smaller than the smallest rectangle containing it (31.5).
>>8121572
You can get a nicer upperbound if you take away the triangle that lowerbounds the empty space.
>>8121572
did you get out your protractor to make sure its a right angle?
>>8121346
This is literally maths for 12 year olds
(4.5*7*Pi)/4
no.
is that curve a segment of a parabola? of an ellipse? of a hyperbola?
we may never know for sure
Model it as the quadratic (7/20.25)(x+4.5)(x-4.5) and integrate between x=0 and 4.5 to get an approximation. If it's an elipse idk
We can only assume that it is to scale.
Measure the area using a grid. If the vertical length is not proportionate to the horizontal length, then the vertical length is at a linear or logarithmic scale to the horizontal length. From here, we can calculate the area.
