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

The correct answer is 36
$y^2-2y=-x$
$\Rightarrow y^2-2y+1=-x+1$
$(y-1{)}^2=-(x-1)$
$y=-x$
Points of intersection
$x^2+2x=-x$
$x^2+3x=0$
$x=0,-3$

$A=\underset{0}{\overset{3}{\int }}\left(-y^2+2y+y\right)dy$
$=\frac{3y^2}{2}-{\frac{y^3}{3}|}_0^3=\frac{9}{2}$
$8A=36$

Answer 2

Step 1: Rewrite the equations in standard form. 

The given curves are:

\[ y^2 - 2y = -x \quad \text{and} \quad x + y = 0. \]

From the second equation, we get:

\[ x = -y. \]

Substituting \(x = -y\) into the first equation:

\[ y^2 - 2y = -(-y) \quad \Rightarrow \quad y^2 - 3y = 0 \quad \Rightarrow \quad y(y - 3) = 0. \]

Thus, \( y = 0 \) and \( y = 3 \).

Step 2: Find the points of intersection.

Using \( y = 0 \) and \( y = 3 \), substitute back into \( x = -y \):

• When \( y = 0 \), \( x = 0 \);
• When \( y = 3 \), \( x = -3 \).

Thus, the points of intersection are \( (0, 0) \) and \( (-3, 3) \).

Step 3: Set up the integral for the area.

The area \( A \) is given by the integral:

\[ A = \int_{y=0}^{y=3} \left( -y - (-y^2 + 2y) \right) \, dy, \] where \( -y \) is the x-coordinate of the line and \( -y^2 + 2y \) is the x-coordinate of the parabola.

Step 4: Simplify the integrand.

\[ A = \int_{y=0}^{y=3} \left( -y + y^2 - 2y \right) \, dy = \int_{y=0}^{y=3} \left( y^2 - 3y \right) \, dy. \]

Step 5: Compute the integral.

\[ A = \int_{y=0}^{y=3} y^2 \, dy - \int_{y=0}^{y=3} 3y \, dy. \] Evaluate each term: \[ \int_{y=0}^{y=3} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{3} = \frac{27}{3} - 0 = 9, \] and \[ \int_{y=0}^{y=3} 3y \, dy = \left[ \frac{3y^2}{2} \right]_{0}^{3} = \frac{27}{2} - 0 = \frac{27}{2}. \] Thus: \[ A = 9 - \frac{27}{2} = \frac{18}{2} - \frac{27}{2} = \frac{9}{2}. \]

Step 6: Calculate \( 8A \).

\[ 8A = 8 \times \frac{9}{2} = 36. \]