Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle. x = 4 − y2, x = y2 − 4

Answer 1

The region enclosed by the given curve is integrated with respect to y and the area is 21.33 square units.

In this question,

The curves are x = 4 - y^2 -------- (1) and

x = y^2 - 4 ------- (2)

The limits of the integral can be found by solving these two curves simultaneously.

On equating (1) and (2),

$4 - y^2 = y^2 - 4$

⇒ $4 +4 = y^2 +y^2$

⇒ $8= 2y^2$

⇒ $y^2=\frac{8}{2}$

⇒ $y^2=4$

⇒ y = +2 or -2

The limits of y is {-2 < y +2} or 2{0 < y < 2}

The diagram below shows the region enclosed by the two curves.

The region enclosed by the given curves can be integrated with respect to y as

$A=2\int\limits^2_0 {[(4-y^{2})-(y^{2}-4 )] } \, dy$

⇒ $A=2\int\limits^2_0 {[4-y^{2}-y^{2}+4 ] } \, dy$

⇒ $A=2\int\limits^2_0 {[8-2y^{2} ] } \, dy$

⇒ $A=2[8y-\frac{2y^{3} }{3} ]\limits^2_0$

⇒ $A=2[8(2)-\frac{2(2)^{3} }{3} ]$

⇒ $A=2[16-\frac{16}{3} ]$

⇒ $A=2[\frac{48-16}{3} ]$

⇒ $A=2[\frac{32}{3} ]$

⇒ $A=\frac{64}{3}$

⇒ $A=21.33$

Hence we can conclude that the region enclosed by the given curve is integrated with respect to y and the area is 21.33 square units.

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