# Find c > 0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -...

## Question:

Find c > 0 such that the area of the region enclosed by the parabolas {eq}y = x^2 - c^2 {/eq} and {eq}y = c^2 - x^2 {/eq} is 210.

## Area bounded by curves:

To find the area bounded by two curves, first of all find the limits of integration by equating both equations, then integrate the function Integration. Area bounded by the two curves {eq}y=f(x) {/eq} and {eq}y=g(x) {/eq} is expressed by the formula {eq}A=\int_{a}^{b} [f(x)-g(x)]dx {/eq}, where a and b are limits of integration.

The given region are

{eq}y=x^2-c^2 {/eq} and {eq}y=c^2-x^2 {/eq}

Equate both the equations

Thus {eq}c^2-x^2=x^2-c^2 {/eq}

{eq}c^2=x^2 {/eq}

{eq}x=\pm c {/eq}

Thus limits of integration for x are from {eq}-c {/eq} to {eq}c {/eq}

Area between the curves

{eq}A=\int_{-c}^{c} [ (c^2-x^2)-(x^2-c^2) ]dx {/eq}

{eq}A=\int_{-c}^{c} (2c^2-2x^2)dx {/eq}

{eq}A=[ 2c^2x-\frac{2x^3}{3} ]_{-c}^{c} {/eq}

As area given is {eq}A=210 {/eq}

{eq}210=[ 2c^3+2c^3-\frac{2c^3}{3}-\frac{2c^3}{3} ] {/eq}

{eq}210= \frac{8c^3}{3} {/eq}

{eq}c^3=78.75 {/eq}

{eq}c= 4.22 {/eq}