Question

Find the area of the region bounded by x 2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Answer 1

Answer:

(32 - 8√2)/3 sq units

Step-by-step explanation:

x² = 4y

=> x = 2√y

=> x = 2 (y)^(1/2)

Integrating with y = 2 & y = 4 boundaries

= 2 *  y ^(3/2) / (3/2)   + c

at y = 2  

= 2 * 2^(3/2) / (3/2)  + c

= (4/3) * 2 √2 + c

= 8√2 / 3 + c

at y = 4

= 2 * 4^(3/2) / (3/2)  + c

= (4/3) * 8 + c

= 32 / 3 + c

Area = | 32 / 3 + c - (8√2 / 3 + c) |

=> Area = | (32 - 8√2)/3 |

Area = (32 - 8√2)/3 sq units

Answer 2

Answer:

Step-by-step explanation:

x² = 4y

=> x = 2√y

=> x = 2 (y)^(1/2)

Integrating with y = 2 & y = 4 boundaries

= 2 * y ^(3/2) / (3/2) + c

at y = 2

= 2 * 2^(3/2) / (3/2) + c

= (4/3) * 2 √2 + c

= 8√2 / 3 + c

at y = 4

= 2 * 4^(3/2) / (3/2) + c

= (4/3) * 8 + c

= 32 / 3 + c

Area = | 32 / 3 + c - (8√2 / 3 + c) |

=> Area = | (32 - 8√2)/3 |