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4 years ago

Consider the region bounded by 4y=x^2 and 2y=x.

Mathematics

1 answer:

gayaneshka [121]4 years ago

7 0

Answer:

a) ⅓ units²

b) 4/15 pi units³

c) 2/3 pi units³

Step-by-step explanation:

4y = x²

2y = x

4y = (2y)²

4y = 4y²

4y² - 4y = 0

y(y-1) = 0

y = 0, 1

x = 0, 2

Area

Integrate: x²/4 - x/2

From 0 to 2

(x³/12 - x²/4)

(8/12 - 4/4) - 0

= -⅓

Area = ⅓

Volume:

Squares and then integrate

Integrate: [x²/4]² - [x/2]²

Integrate: x⁴/16 - x²/4

x⁵/80 - x³/12

Limits 0 to 2

(2⁵/80 - 2³/12) - 0

-4/15

Volume = 4/15 pi

About the x-axis

x² = 4y

x² = 4y²

Integrate the difference

Integrate: 4y² - 4y

4y³/3 - 2y²

Limits 0 to 1

(4/3 - 2) - 0

-2/3

Volume = ⅔ pi

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