What is area 4x+y^2=32,x=y ?

The answer to area 4x+y^2=32,x=y is 72

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area 4x+y^2=32,x=y

area

4 x + y^{2} = 32, x = y

72

Solution steps

Isolate

y

for

4 x + y^{2} = 32 : y = 2 8 - x, y = - 2 8 - x

Isolate

y

for

x = y : y = x

f_{1} (x) = 2 8 - x

f_{2} (x) = - 2 8 - x

To find the intersection points solve

f_{i} (x) = f_{j} (x)

2 8 - x = - 2 8 - x : x = 8

2 8 - x = x : x = 4

- 2 8 - x = x : x = - 8

= \int_{- 8}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{4}^{8} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 8}^{4} - 2 8 - x - x d x + \int_{4}^{8} 2 8 - x - (- 2 8 - x) d x

\int_{- 8}^{4} - 2 8 - x - x d x = \frac{152}{3}

\int_{4}^{8} 2 8 - x - (- 2 8 - x) d x = \frac{64}{3}

= \frac{152}{3} + \frac{64}{3}

Simplify

= 72

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