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Question

Mathematics Question on Area between Two Curves

The area bounded by the curve
y=∣x2−9∣y=|x^2−9|
and the line y = 3 is

A

4(23+6−4)4(2\sqrt3+\sqrt6-4)

B

4(43+6−4)4(4\sqrt3+\sqrt6-4)

C

8(43+36−9)8(4\sqrt3+3\sqrt6-9)

D

8(43+6−9)8(4\sqrt3+\sqrt6-9)

Answer

4(23+6−4)4(2\sqrt3+\sqrt6-4)

Explanation

Solution

The correct asnwer is (D) : 8(43+6−9)8(4\sqrt3+\sqrt6-9)

Fig.

=2∫03(9+y−9−y)dy+2∫39(9−y)dy=2\int^{3}_{0}(\sqrt{9+y}-\sqrt{9-y})dy + 2\int^{9}_{3}(\sqrt{9-y})dy
=2∫03((9+y)1/2−(9−y)dy+∫39(9−y)1/2)dy=2\int^{3}_{0}((9+y)^{1/2}-(9-y)dy + \int^{9}_{3}({9-y})^{1/2})dy
=2[23[(9+y)3/2]03+23[(9−y)3/2]03−23[(9−y)3/2]03]=2[\frac{2}{3}[(9+y)^{3/2}]^{3}_{0} + \frac{2}{3}[(9-y)^{3/2}]^{3}_{0}-\frac{2}{3}[(9-y)^{3/2}]^{3}_{0}]
=43[1212−27+66−27−(0−66)]= \frac{4}{3}[12\sqrt{12}-27+6\sqrt6-27-(0-6\sqrt6)]
=43[243+126−54]=\frac{4}{3}[24\sqrt3+12\sqrt6-54]
=8(43+26−9)=8(4\sqrt3+2\sqrt6-9)