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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),
[tex]4 - y^2 = y^2 - 4[/tex]
⇒ [tex]4 +4 = y^2 +y^2[/tex]
⇒ [tex]8= 2y^2[/tex]
⇒ [tex]y^2=\frac{8}{2}[/tex]
⇒ [tex]y^2=4[/tex]
⇒ 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
[tex]A=2\int\limits^2_0 {[(4-y^{2})-(y^{2}-4 )] } \, dy[/tex]
⇒ [tex]A=2\int\limits^2_0 {[4-y^{2}-y^{2}+4 ] } \, dy[/tex]
⇒ [tex]A=2\int\limits^2_0 {[8-2y^{2} ] } \, dy[/tex]
⇒ [tex]A=2[8y-\frac{2y^{3} }{3} ]\limits^2_0[/tex]
⇒ [tex]A=2[8(2)-\frac{2(2)^{3} }{3} ][/tex]
⇒ [tex]A=2[16-\frac{16}{3} ][/tex]
⇒ [tex]A=2[\frac{48-16}{3} ][/tex]
⇒ [tex]A=2[\frac{32}{3} ][/tex]
⇒ [tex]A=\frac{64}{3}[/tex]
⇒ [tex]A=21.33[/tex]
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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