Answer to Question #176650 in Calculus for Joshua

Question

Find the area bounded by the curve y^2-3x+3=0 and the line x=4.

Answer 1

We can consider the sketch of the graph below

Substituting x=4 in the equation $y^2-3x+3=0$

We get $y^2-3(4)+3=0$

$y^2- 9=0$

$y^2=9$

$y=-3\space or +3$

Intersection points are A(4,-3) and B(4,3)

y is changing from -3 to 3

$Area=\int_a^b(Right \space curve-Left \space curve)dy$

$Area=\int_{-3}^3(4-(1+ \frac{y^2}{3}))dy$

$Area=\int_{-3}^3(4-1- \frac{y^2}{3})dy$

$Area=\int_{-3}^3(3- \frac{y^2}{3})dy$

$Area=\int_{-3}^33dy- \int_{-3}^3\frac{y^2}{3}dy$

$Area=3[y]_{-3}^3-\frac{1}{3}[\frac{y^3}{3}]_{-3}^3$

$Area=18-6=12$ square units