Find the values of #c# such that the area...?

Question

Find the values of #c# such that the area of the region bounded by the parabolas #y=x^2-c^2# and #y=c^2-x^2# is 576.

Emilia Aguilar · Accepted Answer

#c =6#

The two curves are:

#y_1(x) = x^2-c^2#

and

#y_2(x) = c^2-x^2#

We can note that for every #x# we have: #y_1(x) = -y_2(x)# so the two parabolas are symmetric with respect to the #x# axis.The two curves thus intercept when #y_1(x) = y_2(x) = 0#, that is for #x=+-c#

Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola and the #x# axis.

If we choose #y_2(x) = c^2-x^2#, which is positive in the interval, we thus have:

#A = 2 int_(-c)^c ( c^2-x^2)dx = 2 [c^2x-x^3/3]_(-c)^c = 8/3c^3#

and posing

#8/3c^3 = 576#

we get:

#c = root(3)((3 xx 576)/8) = 6#

Ezra Adams · Answer

undefined To find the value of $ c $ such that the area enclosed by the curves $ y = x^2 $ and $ y = c - x^2 $ is $ 16 $ square units, we set up the definite integral $$ \int_{a}^{b} (c - x^2 - x^2) \, dx = 16 $$ and solve for $ c $. The limits of integration $ a $ and $ b $ are the points of intersection of the two curves. Solving for $ a $ and $ b $, we find $ a = -\sqrt{c} $ and $ b = \sqrt{c} $. Substituting these values into the integral and solving for $ c $, we get $ c = 8 $. Thus, the value of $ c $ such that the area enclosed by the curves is $ 16 $ square units is $ c = 8 $.