Find the area enclosed by the curves x = 0, -x = y^2, and 2x + 2y + 4 = 0
- Thread starter heyitsme
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To follow up, we want to find where the non-vertical line and the parabola intersect, so let's first simplify the line:
[MATH]x+y+2=0[/MATH]
Substitute for \(x\) from the parabola:
[MATH]-y^2+y+2=0[/MATH]
Or:
[MATH]y^2-y-2=0[/MATH]
Factor:
[MATH](y-2)(y+1)=0[/MATH]
[MATH]y=-1\implies x=-1[/MATH]
[MATH]y=2\implies x=-4[/MATH]
Similarly, we find the vertical line and the non-vertical line intersect at \((0,-2)\) and the vertical line and the parabola intersect at \((0,0)\). Hence, there are two ways we can compute the bounded area \(A\):
1.) [MATH]A=\int_{-1}^0 -\sqrt{-x}-(-(x+2))\,dx[/MATH]
Let:
[MATH]u=-x\implies du=-dx[/MATH]
And we have:
[MATH]A=\int_0^1 2-u-\sqrt{u}\,du=\frac{1}{6}\left[12u-3u^2-4u^{\frac{3}{2}}\right]_0^1=\frac{5}{6}[/MATH]
2.) [MATH]A=\int_{-2}^{-1} 0-(-(y+2))\,dx+\int_{-1}^0 0-(-y^2)\,dy [/MATH]
[MATH]A=\frac{1}{2}\left[y^2+4y\right]_{-2}^{-1}+\frac{1}{3}\left[y^3\right]_{-1}^0=\frac{1}{2}(-3+4)+\frac{1}{3}(0+1)=\frac{5}{6}[/MATH]
[MATH]x+y+2=0[/MATH]
Substitute for \(x\) from the parabola:
[MATH]-y^2+y+2=0[/MATH]
Or:
[MATH]y^2-y-2=0[/MATH]
Factor:
[MATH](y-2)(y+1)=0[/MATH]
[MATH]y=-1\implies x=-1[/MATH]
[MATH]y=2\implies x=-4[/MATH]
Similarly, we find the vertical line and the non-vertical line intersect at \((0,-2)\) and the vertical line and the parabola intersect at \((0,0)\). Hence, there are two ways we can compute the bounded area \(A\):
1.) [MATH]A=\int_{-1}^0 -\sqrt{-x}-(-(x+2))\,dx[/MATH]
Let:
[MATH]u=-x\implies du=-dx[/MATH]
And we have:
[MATH]A=\int_0^1 2-u-\sqrt{u}\,du=\frac{1}{6}\left[12u-3u^2-4u^{\frac{3}{2}}\right]_0^1=\frac{5}{6}[/MATH]
2.) [MATH]A=\int_{-2}^{-1} 0-(-(y+2))\,dx+\int_{-1}^0 0-(-y^2)\,dy [/MATH]
[MATH]A=\frac{1}{2}\left[y^2+4y\right]_{-2}^{-1}+\frac{1}{3}\left[y^3\right]_{-1}^0=\frac{1}{2}(-3+4)+\frac{1}{3}(0+1)=\frac{5}{6}[/MATH]