Help with problem

Can someone help me figure out the area for this please?

Yes

>ft

no

double integral

dy from x^2 to sqrt(7)
dx from 0 to 4.5

Is "ft" the unit or is something else. Because if it's the unit (feet) the area can't be calculated

so hows that curve defined? there exists an infinite number of circles going through two points

It's obviously not a circle..

So the vertical distance is 20 ft?

if it's so obvious, what is it then?

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

= 0 rather

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.

False dichotomy mate :)
It's obvious what it isn't, but not what it is.

I like you

Feet is a unit of area now?

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).

You can get a nicer upperbound if you take away the triangle that lowerbounds the empty space.

did you get out your protractor to make sure its a right angle?

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.