Find the area of the region enclosed by curves $2 x + y^{2} = 48$ and $x = y$ .

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Solution

$\begin{matrix} w e h a v e t h e g i v e n e q u a t i o n 2 x + y^{2} = 48 a n d x = y . s o l v i n g f o r x f r o m t h e g i v e n e q u a t i n w e g e t x = - \frac{1}{2} y^{2} + 24 x = y T o f i n d t h e p o i n t o f i n t e r sec t i o n w e h a v e y = - \frac{1}{2} y^{2} + 24 2 y = - y^{2} + 48 y^{2} + 2 y - 48 = 0 (y + 8) (y - 6) = 0 S o, t h e p o i n t s o f i n t e r s e c t i o n a r e w h e n x = - 8, 6. s i n c e, t h e 2 x + y^{2} = 48 f u n c t i o n i s t h e r i g h t b o u n d e d w e h a v e a n a r e a o f \int_{- 8}^{6} (- \frac{1}{2} y^{2} + 24) - y d y = {[- \frac{1}{6} y^{3} + 24 y - \frac{1}{2} y^{2}]}_{- 8}^{6} = - \frac{1}{6} {(6)}^{3} + 24 (6) - \frac{1}{2} {(6)}^{2} - (- \frac{1}{2} {(- 8)}^{3} + 24 (- 8) - \frac{1}{2} {(- 8)}^{2}) = 90 - (\frac{- 416}{3}) = \frac{686}{3} \end{matrix}$

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