Question

How do you find the area between $ x = 4 - {y^2} $ and $ x = y - 2 $ ?

Answer 1

Hint: In order to find the area, we need to know the region between the equations, and from the equation we can see that the first equation would represent a parabola and the second one would represent a straight line. Compare the two equations to find the points of intersection then using integration find the area inside them.
Formula used:
 $ \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
 $ \int {adx = } a\int {dx} $

Complete step-by-step answer:
The first Equation given is: $ x = 4 - {y^2} $ and the second is $ x = y - 2 $ .
Since, value of $ x $ is given in both, so comparing the two equations, we get:
 $
  4 - {y^2} = y - 2 \\
  {y^2} + y - 6 = 0 \;
  $
Comparing the obtained Quadratic Equation with the standard Quadratic Equation $ a{x^2} + bx + c = 0 $ , we get:
 $
  a = 1 \\
  b = 1 \\
  c = - 6 \;
  $
Solving for discriminant, we get:
 $
  D = \sqrt {{b^2} - 4ac} \\
  D = \sqrt {{1^2} - 4 \times 1 \times \left( { - 6} \right)} \\
  D = \sqrt {1 + 24} \\
  D = \sqrt {25} = 5 \;
  $
Quadratic Formula to find both roots of a quadratic equation as
 $ y_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} $ and $ y_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} $
 $ y_1,y_2 $ are root to quadratic equations $ a{x^2} + bx + c $ .
For, $ y_1 $ :
 $
  y_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \\
  y_1 = \dfrac{{ - 1 + 5}}{{2 \times 1}} = \dfrac{4}{2} = 2 \\
  $
For, $ y_2 $ :
 $
  y_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\
  y_2 = \dfrac{{ - 1 - 5}}{{2 \times 1}} = \dfrac{{ - 6}}{2} = - 3 \;
  $
 Hence the factors will be $ (y - y_1)\,and\,(y - y_2)\, $ that is $ (y - 2)\,and\,(y + 3)\, $ .
Putting the values of $ y_1 $ and $ y_2 $ in the equation $ x = y - 2 $ one by one to get $ x_1 $ and $ x_2 $ :
 $ x_1 = y_1 - 2 = 2 - 2 = 0 $
 $ x_2 = y_2 - 2 = - 3 - 2 = - 5 $
Therefore, the points of intersection are:
 $ (x_1,y_1)\,and\,(x_2,y_2)\, $ , which are $ (0,2)\,and\,( - 5, - 3)\, $ .
According to the points obtained, the graph would be:

Since, the starting and ending points for the y-axis are: $ - 3and2 $ .So, the area is starting from $ - 3to2 $ .
Writing the first and second equation in terms of $ y $ and we get:
 $ x = 4 - {y^2} = > y = \sqrt {4 - x} $
 $ x = y - 2 = > y = x + 2 $
The area under the equation is:
 $ A = \int\limits_{ - 3}^2 {\left[ {\left( {4 - {y^2}} \right) - \left( {y - 2} \right)} \right]} dy $
On further solving, we get:
 $
  A = \int\limits_{ - 3}^2 {\left[ {\left( {4 - {y^2}} \right) - \left( {y - 2} \right)} \right]} dy \\
  A = \int\limits_{ - 3}^2 {\left[ {4 - {y^2} - y + 2} \right]} dy \\
  A = \int\limits_{ - 3}^2 {\left[ { - {y^2} - y + 6} \right]} dy \\
  A = - \int\limits_{ - 3}^2 {{y^2}} dy - \int\limits_{ - 3}^2 y dy + 6\int\limits_{ - 3}^2 {dy} \\
  A = - {\left[ {\dfrac{{{y^3}}}{3}} \right]_{ - 3}}^2 - {\left[ {\dfrac{{{y^2}}}{2}} \right]^2}_{ - 3} + 6{\left[ y \right]^2}_{ - 3} \\
  A = - \left[ {\dfrac{{{2^3}}}{3} - \dfrac{{{{\left( { - 3} \right)}^3}}}{3}} \right] - \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{{\left( { - 3} \right)}^2}}}{2}} \right] + 6\left[ {2 - \left( { - 3} \right)} \right] \\
  A = - \left[ {\dfrac{8}{3} + 9} \right] - \left[ {2 - \dfrac{9}{2}} \right] + 6\left[ {2 + 3} \right] \\
  A = - \dfrac{8}{3} - 9 - 2 + \dfrac{9}{2} + 30 \\
  A = - \dfrac{8}{3} - 11 + \dfrac{9}{2} + 30 \\
  A = \dfrac{{125}}{6} = 20.833 \;
  $
Therefore, the area between $ x = 4 - {y^2} $ and $ x = y - 2 $ is $ 20.833 $ sq. units.
So, the correct answer is “ $ 20.833 $ sq. units.”.

Note: It's important to find the intersecting points to know the area covered by the parabola and the straight line.
We can also take the value of $ x $ instead of y to find the area, just the values inside the integration would be written in terms of $ x $ and $ dx $ .