Find the area of the region bounded by the graphs x=sqrt 2sin2y ; x=0
x=sqrt 2sin2y ; x=0
To find the area of the region bounded by the graphs, we can use the method of integration.
Step 1: Determine the limits of integration.
In this problem, we are given that x = 0 bounds the region, and we need to find the limits for y. To do this, we can set the two equations equal to each other and solve for y:
sqrt(2)sin(2y) = 0
sin(2y) = 0
2y = nπ (where n is an integer)
y = nπ/2 (n is an integer)
Since we want to find the area between the x-axis and the curve, we only need to consider the positive values of y. So, the limits of integration for y are from 0 to π/2.
Step 2: Write the integral expression.
The area (A) can be found by integrating the function over the given limits:
A = ∫[0, π/2] sqrt(2)sin(2y) dy
Step 3: Evaluate the integral.
The integral of sqrt(2)sin(2y) can be evaluated using substitution method. Let's substitute u = 2y, then du = 2dy. The bounds of integration will change accordingly.
A = ∫[0, π/2] (1/2) sqrt(2)sin(u) du
A = (1/2) ∫[0, π] sqrt(2)sin(u) du
Using the trigonometric identity sin(u) = sin(π - u), we can rewrite the integral as follows:
A = (1/2) ∫[0, π] sqrt(2)sin(π - u) du
Now, let's evaluate the integral:
A = (1/2) [-sqrt(2)cos(u)] [0, π]
A = (1/2) [-sqrt(2)cos(π) + sqrt(2)cos(0)]
A = (1/2) [-sqrt(2)(-1) + sqrt(2)(1)]
A = sqrt(2)
Therefore, the area of the region bounded by the graphs x = sqrt(2)sin(2y) and x = 0 is sqrt(2) square units.
To find the area of the region bounded by the graphs x = √(2sin(2y)) and x = 0, we need to set up an integral.
First, let's find the limits of integration. Looking at the given graphs, we can see that x ranges from 0 to √(2sin(2y)). To find the corresponding limits for y, we set the two equations equal to each other:
√(2sin(2y)) = 0
But since the square root of a non-negative number is always non-negative, we can conclude that the given region is bounded by y, where y ranges from 0 to π/2.
Now, let's set up the integral to calculate the area. The differential form of area is given by dA = f(x)dx, where f(x) is the difference in y-coordinates at any given x.
Since x is bounded by the graphs x = √(2sin(2y)) and x = 0, and y is bounded by 0 and π/2, the area can be expressed as:
A = ∫[0,π/2] √(2sin(2y)) dx
But dx can be expressed in terms of dy using the chain rule:
dx = (√(2sin(2y)))' dy
Taking the derivative of √(2sin(2y)) with respect to y:
dx = (√(2sin(2y)))' dy
= (√(2sin(2y))) * (2cos(2y)) * 2 dy
= 4cos(2y)√(2sin(2y)) dy
Therefore, the area becomes:
A = ∫[0,π/2] (4cos(2y)√(2sin(2y)) dy)
Integrating this expression within the given limits will give you the area of the region bounded by the graphs x = √(2sin(2y)) and x = 0.