Calculate the area inside both of the ellipses (x^2)/3 + y^2 =1 and x^2 + (y^2)/3 = 1.
To calculate the area inside both of the ellipses, we can use the concept of double integration in polar coordinates.
Step 1: Convert the equations of the ellipses into polar coordinates.
For the first ellipse, we have:
(x^2)/3 + y^2 = 1
Converting this equation into polar coordinates, we use the following relationships:
x = r * cos(theta)
y = r * sin(theta)
Simplifying the equation, we have:
(r^2 * cos^2(theta))/3 + (r^2 * sin^2(theta)) = 1
r^2 * [(cos^2(theta))/3 + (sin^2(theta))] = 1
r^2 = 3/(cos^2(theta) + 3sin^2(theta))
Similarly, for the second ellipse, we have:
x^2 + (y^2)/3 = 1
Converting into polar coordinates, we get:
r^2 * [(cos^2(theta)) + (sin^2(theta))/3] = 1
r^2 = 1/[(cos^2(theta)) + (sin^2(theta))/3]
Step 2: Determine the limits of integration for theta and r.
To find the limits of integration for theta, we need to determine the angles at which the ellipses intersect.
Since the equations are in standard form, we can see that the ellipses intersect when:
(cos^2(theta))/3 + sin^2(theta) = cos^2(theta) + (sin^2(theta))/3
Multiplying through by 3, we get:
cos^2(theta) + 3sin^2(theta) = 3cos^2(theta) + sin^2(theta)
Combining like terms, we have:
2sin^2(theta) = 2cos^2(theta)
sin^2(theta) = cos^2(theta)
Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1, we can substitute:
1 - cos^2(theta) = cos^2(theta)
1 = 2cos^2(theta)
cos^2(theta) = 1/2
Taking the square root:
cos(theta) = ±(1/√2)
Since cosine is positive in the first and fourth quadrants, we consider the positive value:
cos(theta) = 1/√2
Taking the arccosine:
theta = π/4
Thus, the limits of integration for theta are 0 to π/4.
For r, we need to find the maximum and minimum values of r for each ellipse. The maximum radius can be found by taking the square root of the reciprocal of the smaller coefficient in each equation.
For the first ellipse, the smaller coefficient is 1/3:
r_max1 = √(1/(1/3)) = √3
For the second ellipse, the smaller coefficient is 1/3:
r_max2 = √(1/(1/3)) = √3
Therefore, the limits of integration for r are 0 to √3 for both ellipses.
Step 3: Set up the double integral to calculate the area.
The area inside both ellipses can be calculated by setting up a double integral in polar coordinates. The double integral integrates r from 0 to √3 and theta from 0 to π/4.
Area = ∫∫ r dr dθ (from 0 to √3 and 0 to π/4)
Step 4: Evaluate the double integral to find the area.
Integrating with respect to r first, we have:
∫[0 to π/4] ∫[0 to √3] r dr dθ
Integrating with respect to r, we get:
(1/2) r^2 [from 0 to √3] = (1/2) (√3)^2 - (1/2) (0)^2 = (1/2) * 3 = 3/2
Finally, evaluating the remaining integral with respect to theta:
∫[0 to π/4] (3/2) dθ = (3/2) θ [from 0 to π/4] = (3/2) (π/4 - 0) = 3π/8
Therefore, the area inside both ellipses is 3π/8 square units.