38.4k views
0 votes
find the area of the curtain wall below the graph of z = 4y , y ≥ 0 , and above the parabola y 2 = x , 0 ≤ x ≤ 2 , in the xy-plane.

User Taty
by
8.1k points

1 Answer

4 votes

Final answer:

To find the area of the curtain wall below the graph of z = 4y, y ≥ 0, and above the parabola y^2 = x, 0 ≤ x ≤ 2, we can find the intersection points of the two curves and calculate the area between them. The intersection points are y = 0 and y = 4. By integrating the function (4y - y^2) with respect to y from y = 0 to y = 4, we find that the area of the curtain wall is 32 - 64/3 square units.

Step-by-step explanation:

To find the area of the curtain wall below the graph of z = 4y, y ≥ 0, and above the parabola y^2 = x, 0 ≤ x ≤ 2, we need to find the intersection points of the two curves and calculate the area between them.

To find the intersection points, we set z = 4y equal to y^2 = x.

Substituting z = 4y into y^2 = x, we get:

4y^2 = x

Now we can find the intersection points by solving the equation:

y^2 = x = 4y

y^2 - 4y = 0

y(y - 4) = 0

So the intersection points are y = 0 and y = 4.

Now we can calculate the area between the curves using the formula for finding the area between two curves:

Area = ∫(4y - y^2) dx from x = 0 to x = 2

Area = ∫(4y - y^2) dx from x = 0 to x = 2

Since the area is between y = 0 and y = 4, we integrate with respect to y:

Area = ∫(4y - y^2) dy from y = 0 to y = 4

Integrating, we get:

Area = [2y^2 - (y^3/3)] from y = 0 to y = 4

Area = 2(4^2) - (4^3/3) - 0 = 32 - 64/3

So the area of the curtain wall is 32 - 64/3 square units.

User Charles Miller
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories