Question

If the area of the region bounded by the curves $y ^2-2 y=-x, x+y=0$ is $A$, then $8 A$ is equal to _________

Answer 1

Correct Answer: 36

To determine the area \( A \) such that \( 8A \) is the answer, we first analyze the curves: \( y^2-2y=-x \) and \( x+y=0 \).

Rewrite \( x+y=0 \) as \( x=-y \). Substitute \( x=-y \) into the first curve:

\( y^2-2y=y \)

\( y^2-3y=0 \)

\( y(y-3)=0 \)

This gives the solutions \( y=0 \) or \( y=3 \). For these values of \( y \), \( x=0 \) when \( y=0 \) and \( x=-3 \) when \( y=3 \).

Next, solve \( y^2-2y=-x \) using \( x=-y \):

\( y^2-2y+y=0 \)

\( y(y-1)=0 \)

This gives \( y=0 \) or \( y=1 \). Consequently, when \( y=0 \), \( x=0 \). When \( y=1 \), \( x=-1 \).

Identify the bounded region: the line \( x=-y \) intersects the parabola at points \((0,0)\), and the analysis above gives us \((0,0)\) and \((-3,3)\) as the limits.

Set the integrals with respect to \( y \) between limits 0 and 3 for the area calculation. We find:

The lines intersect at two points: \((0,0)\) and \((-3,3)\). For area \( A \), integrate the difference of \( x \) coordinates:

\(\int_0^3 ((y^2-2y)-(-y)) \, dy = \int_0^3 (y^2-y) \, dy\)

Calculate the definite integral:

\(\int_0^3 y^2\, dy - \int_0^3 y\, dy\)

\(\left[\frac{y^3}{3}\right]_0^3 - \left[\frac{y^2}{2}\right]_0^3\)

\(\left[\frac{27}{3}\right] - \left[\frac{9}{2}\right] = 9 - 4.5 = 4.5\)

The area \( A \) is 4.5. Thus, \( 8A = 8 \times 4.5 = 36\).

Since \( 8A \) is within the range of 36, the solution is correct.