Calculate the area enclosed by the curves x^2 | vedclass

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(C) The given curves are $x^2 = 2 - y$ (which is $y = 2 - x^2$) and $x^2 = y$.To find the intersection points,set $2 - x^2 = x^2$.$2x^2 = 2 \Rightarro

Vedclass · Accepted Answer

$\frac{8}{3}$ square units

Vedclass · Answer

(C) The given curves are $x^2 = 2 - y$ (which is $y = 2 - x^2$) and $x^2 = y$.To find the intersection points,set $2 - x^2 = x^2$.$2x^2 = 2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.$A = \int_{-1}^{1} ((2 - x^2) - x^2) dx = \int_{-1}^{1} (2 - 2x^2) dx$.Since the function is even,$A = 2 \int_{0}^{1} (2 - 2x^2) dx$.$A = 4 \int_{0}^{1} (1 - x^2) dx = 4 [x - \frac{x^3}{3}]_{0}^{1}$.$A = 4 (1 - \frac{1}{3}) = 4 (\frac{2}{3}) = \frac{8}{3}$ square units.

Calculate the area enclosed by the curves $x^2 = 2 - y$ and $x^2 = y$.

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