What is area x=(y-1)^2,x=13-y^2 ?

The answer to area x=(y-1)^2,x=13-y^2 is 125/3

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area x=(y-1)^2,x=13-y^2

area

x = (y - 1)^{2}, x = 13 - y^{2}

\frac{125}{3}

Decimal Notation

41.66666 \dots

Solution steps

Isolate

y

for

x = (y - 1)^{2} : y = x + 1, y = - x + 1

Isolate

y

for

x = 13 - y^{2} : y = - x + 13, y = - - x + 13

f_{1} (x) = x + 1

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

To find the intersection points solve

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

x + 1 = - x + 1 : x = 0

x + 1 = - x + 13 : x = 4

x + 1 = - - x + 13 :

No Solution for

x \in R

- x + 1 = - x + 13 :

No Solution for

x \in R

- x + 1 = - - x + 13 : x = 9

- x + 13 = - - x + 13 : x = 13

= \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{9} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{9}^{13} ∣ f_{3} (x) - f_{4} (x) ∣ d x

= \int_{0}^{4} x + 1 - (- x + 1) d x + \int_{4}^{9} - x + 1 - - x + 13 d x + \int_{9}^{13} - x + 13 - (- - x + 13) d x

\int_{0}^{4} x + 1 - (- x + 1) d x = \frac{32}{3}

\int_{4}^{9} - x + 1 - - x + 13 d x = \frac{61}{3}

\int_{9}^{13} - x + 13 - (- - x + 13) d x = \frac{32}{3}

= \frac{32}{3} + \frac{61}{3} + \frac{32}{3}

Simplify

= \frac{125}{3}

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What is area x=(y-1)^2,x=13-y^2 ?
The answer to area x=(y-1)^2,x=13-y^2 is 125/3