Question

Find the value of c for which the area enclosed by the curves y=x^2-c^2 and y=c^2-x^2 is 576?

please do this question.
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Answer 1

`c = 6`

Explanation:

The two curves intersect at the zeros, `(-c,0)` and `(c,0)`; please observe this in the graph were `c = 1`:

graph{(x^2-1-y)(1-x^2-y)=0 [-10, 10, -5, 5]}

Also, please notice that the curve `c^2-x^2 > x^2-c^2` in that interval, therefore, the Area can be calculate using the following equation:

`"Area" = int_-c^c (c^2-x^2)- (x^2-c^2)dx`

This simplifies to:

`"Area" = int_-c^c 2c^2-2x^2dx = 2(c^2x-1/3x^3]_-c^c`

Substitute 576 for the area:

`576 = 2(c^3-1/3c^3- (-c^3+1/3c^3))`

`576 = 2(2c^3-2/3c^3)`

`576 = 8/3c^3`

`c^3 = 216`

`c= 6`

Answer 2

`c=6`

Explanation:

Observe that the curves `y=x^2-c^2` and `y=c^2-x^2` are symmetric w.r.t. `y`-axis and each of the function is reflection of the other in `x`-axis.

As such the area bounded by the two curves is equally divided by the two axes in four quadrants, and area bounded is hence four times the area bounded (in the `Q1`) by `y=c^2-x^2` and `y=0` and `x=0`. (See the graph below, not drawn to scale.)

This area is given by

`int_0^c(c^2-x^2)dx`

= `[c^2x-x^3/3]_0^c`

= `c^3-c^3/3`

= `(2c^3)/3`

and area bounded by two curves is `(8c^3)/3` and as

`(8c^3)/3=576`

`c^3=576xx3/8=72xx3=216` i.e. `c=6`

graph{(y-36+x^2)(y+36-x^2)=0 [-10, 10, -60, 60]}