The area of the region bounded by the parabol | vedclass

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(A) To find the area bounded by the parabola $y^2 = 8x$ and the line $y = -x$,we first find the points of intersection.Substituting $y = -x$ into $y^2

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$\frac{32}{3}$

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(A) To find the area bounded by the parabola $y^2 = 8x$ and the line $y = -x$,we first find the points of intersection.Substituting $y = -x$ into $y^2 = 8x$,we get $(-x)^2 = 8x$,which implies $x^2 - 8x = 0$.Thus,$x(x - 8) = 0$,so $x = 0$ and $x = 8$.For $x = 0$,$y = 0$. For $x = 8$,$y = -8$.The points of intersection are $(0, 0)$ and $(8, -8)$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = 8$.Here,the parabola $y = \pm \sqrt{8x}$ is the upper curve for $y > 0$ and the line $y = -x$ is the lower curve.However,since the region is bounded by $y^2 = 8x$ and $y = -x$,we integrate with respect to $y$ or $x$.Using $x$ as the variable: $A = \int_{0}^{8} (\sqrt{8x} - (-x)) dx = \int_{0}^{8} (2\sqrt{2}x^{1/2} + x) dx$.$A = [2\sqrt{2} \cdot \frac{2}{3} x^{3/2} + \frac{x^2}{2}]_{0}^{8}$.$A = [\frac{4\sqrt{2}}{3} (8^{3/2}) + \frac{64}{2}] = [\frac{4\sqrt{2}}{3} (16\sqrt{2}) + 32] = [\frac{64 \cdot 2}{3} + 32] = [\frac{128}{3} + 32] = \frac{128 + 96}{3} = \frac{224}{3}$.Wait,re-evaluating the region: The parabola $y^2=8x$ opens to the right. The line $y=-x$ passes through the origin. The area enclosed is $\int_{-8}^{0} ((-y) - (y^2/8)) dy = [-\frac{y^2}{2} - \frac{y^3}{24}]_{-8}^{0} = 0 - (- \frac{64}{2} - \frac{-512}{24}) = -(-32 + \frac{64}{3}) = 32 - \frac{64}{3} = \frac{96-64}{3} = \frac{32}{3}$.

The area of the region bounded by the parabola $y^2 = 8x$ and the line $x + y = 0$ is . . . . . . sq. units.

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