Respuesta :
solution:
[tex]the vartical element has hight equal to the difference between the parabola and x-axis with width dx.\\
\frac{A}{4}=\int [(upper function)-(lower function)]dx\\
upper function y=c^2-16x^2\\
lower function y=0\\[/tex]
[tex]\frac{A}{4}=\int (c^2-16x^2)dx\\
integrate from zero to \frac{c}{4} and remember, A=18\\
\frac{18}{4}=c^2x-(\frac{16}{3})x^3,evaluated from zero to \frac{c}{4}\\
\frac{18}{4}=c^2(\frac{c}{4})-(\frac{16}{3})\int \int (\frac{c^3}{64})\\
\frac{18}{4}=\frac{c^3}{4}-\frac{c^3}{12}\\
\frac{18}{4}=\frac{3c^3}{12}-\frac{c^3}{12}\\
\frac{18}{4}=\frac{c^3}{6}\\
\frac{108}{4}=c^3\\
27=c^3\\
c=3\\[/tex]
[tex]we eveluated the portion of the area lying in the second quadrant, c whould have been -3\\
c=3,-3\\
To check, evaluate the area of half the region, integrating a vertical slide between 9-16x^2 and y=16x^2-9\\
\frac{A}{2}=\int [(upper function)-(lower function)]dx\\
evaluated from zero to \frac{3}{4}\\
\frac{18}{2}=\int [(9-16x^2)-(16x^2-9)]dx\\
9=\int [18-32x^2]dx\\[/tex]
[tex]9=18x-(\frac{32}{3})x^3,evaluated from zero to \frac{3}{4}\\
9=\frac{54}{4}-\frac{9}{2}\\
9=\frac{27}{2}-\frac{9}{2}\\
9=\frac{27-9}{2}\\
9=\frac{18}{2}[/tex]
