Question

Find the values of c such that the area of the region bounded by the parabolas . Y = 16x2 − c2 and y = c2 − 16x2. Is 18

Answer 1

`the vartical element has hight equal to the difference between the parabola and x-axis with width dx.\\</p><p>(A)/(4)=\int [(upper function)-(lower function)]dx\\</p><p>upper function y=c^2-16x^2\\</p><p>lower function y=0\\`

`(A)/(4)=\int (c^2-16x^2)dx\\</p><p>integrate from zero to (c)/(4) and remember, A=18\\</p><p>(18)/(4)=c^2x-((16)/(3))x^3,evaluated from zero to (c)/(4)\\</p><p>(18)/(4)=c^2((c)/(4))-((16)/(3))\int \int ((c^3)/(64))\\</p><p>(18)/(4)=(c^3)/(4)-(c^3)/(12)\\</p><p>(18)/(4)=(3c^3)/(12)-(c^3)/(12)\\</p><p>(18)/(4)=(c^3)/(6)\\</p><p>(108)/(4)=c^3\\</p><p>27=c^3\\</p><p>c=3\\`

`we eveluated the portion of the area lying in the second quadrant, c whould have been -3\\</p><p>c=3,-3\\</p><p>To check, evaluate the area of half the region, integrating a vertical slide between 9-16x^2 and y=16x^2-9\\</p><p>(A)/(2)=\int [(upper function)-(lower function)]dx\\</p><p>evaluated from zero to (3)/(4)\\</p><p>(18)/(2)=\int [(9-16x^2)-(16x^2-9)]dx\\</p><p>9=\int [18-32x^2]dx\\`

`9=18x-((32)/(3))x^3,evaluated from zero to (3)/(4)\\</p><p>9=(54)/(4)-(9)/(2)\\</p><p>9=(27)/(2)-(9)/(2)\\</p><p>9=(27-9)/(2)\\</p><p>9=(18)/(2)`