Answer to Question #334919 in Calculus for Quân Jason

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2y = 4sqrt(x), y=5 and 2x + 2y =6.

Answer 1

$$2x+2y=6=>x=-y+3$$

$$2y = 4\sqrt{x}=>x=\dfrac{1}{4}y^2, y\ge0$$

$$-y+3=\dfrac{1}{4}y^2, y\ge0$$

$$y^2+4y-12=0, y\ge0$$

$$(y-2)(y+6)=0, y\ge0$$

Then $y=2.$

$$Area=A=\displaystyle\int_{2}^{5}(\dfrac{1}{4}y^2-(-y+3))dy$$

$$=\bigg[\dfrac{1}{12}y^3+\dfrac{1}{2}y^2-3y\bigg]\begin{matrix} 5 \\ 2 \end{matrix}$$

$$=\dfrac{125}{12}+\dfrac{25}{2}-15-\dfrac{8}{12}-2+6$$

$$=\dfrac{45}{4}({units}^2)=11.25({units}^2)$$

$Area=11.25$ square units.