Area enclosed between the curves |y|=|1-x2| and x2+y2=1, is

Question

Area enclosed between the curves |y|=|1-x2| and x2+y2=1, is (A) (3 π-8/3) (B) (π-8/3) (C) (2 π-8/3) (D) (π+8/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/c0085e9d162db20efc5b86615a96ac79-.png" /> |y|=1-x2, C1: x2+y2=1 C2: y=1-x2 C3: y=x2-1 text Required area =π-4 Δ1 =π-4 ∫01(1-x2) d x =π-(8/3)=(3 π-8/3)

Q. Area enclosed between the curves $∣ y ∣ = ∣ ∣ 1 - x^{2} ∣ ∣$ and $x^{2} + y^{2} = 1$ , is

$\frac{3 π - 8}{3}$

$\frac{π - 8}{3}$

$\frac{2 π - 8}{3}$

$\frac{π + 8}{3}$

$∣ y ∣ = 1 - x^{2},$
$C_{1} : x^{2} + y^{2} = 1$
$C_{2} : y = 1 - x^{2}$
$C_{3} : y = x^{2} - 1$
$Required area = π - 4 Δ_{1}$
$= π - 4 \int_{0}^{1} (1 - x^{2}) d x$
$= π - \frac{8}{3} = \frac{3 π - 8}{3}$

Q. Area enclosed between the curves ∣y∣=∣∣​1−x2∣∣​ and x2+y2=1, is

33π−8​

3π−8​

32π−8​

3π+8​