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

Mathematics

1 answer:

gayaneshka [121]3 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

You might be interested in

a 12-ft ladder leans against a building so that the angle between the ground and the ladder is 74°. how high does the ladder rea

WINSTONCH [101]

Answer:

12 feet

Step-by-step explanation:

Sometimes, a simple drawing is a great way to start! In the attached drawing, the red line represents the ladder, the brown represents the ground, and the black represents the building!

We want to find the side of the triangle from where the ladder meets the building to where the building meets the ground. This can be found using trigonometric identities, namely sine:

$sin\theta=\frac{opposite}{hypotenuse}$