Answer:
To find the area of the regions enclosed by the given curves and lines using double integrals, we'll use the following formula for the area of a region R in the xy-plane:
\[A = \iint_R dA\]
Here, dA represents an infinitesimal area element. In rectangular coordinates (x, y), we can express dA as dx dy. Thus, our double integral will become:
\[A = \iint_R dx dy\]
We'll calculate the area for each of the four cases (a, b, c, d) separately.
(a) Curves \(y = x^2\), \(y = -x^2\), and lines \(x = 1\), \(x = 2\):
To find the limits of integration for x and y, we'll first determine the intersection points of the curves and lines:
1. \(y = x^2\) and \(y = -x^2\) intersect at \(x = 0\) (the origin).
So, the limits of integration are as follows:
\[x: 0 \leq x \leq 1\]
\[y: -x^2 \leq y \leq x^2\]
Now, we can set up the double integral:
\[A = \int_0^1 \int_{-x^2}^{x^2} dy dx\]
(b) Curve \(x = -y^2\), and lines \(y = x - 4\), \(y = -2\), \(y = 2\):
To find the limits of integration for x and y, we'll first determine the intersection points of the curve and lines:
1. \(x = -y^2\) intersects \(y = x - 4\) when \(x - 4 = -y^2\), which gives us \(y = 2\) and \(x = 0\).
So, the limits of integration are as follows:
\[x: 0 \leq x \leq -y^2\]
\[y: -2 \leq y \leq 2\]
Now, we can set up the double integral:
\[A = \int_{-2}^{2} \int_0^{-y^2} dx dy\]
(c) Curve \(y = 5 - x^2\), and line \(y = x + 3\):
To find the limits of integration for x and y, we'll first determine the intersection points of the curve and line:
1. \(y = 5 - x^2\) intersects \(y = x + 3\) when \(5 - x^2 = x + 3\).
Solving for x, we find \(x = \pm 2\).
So, the limits of integration are as follows:
\[x: -2 \leq x \leq 2\]
\[y: x + 3 \leq y \leq 5 - x^2\]
Now, we can set up the double integral:
\[A = \int_{-2}^{2} \int_{x + 3}^{5 - x^2} dy dx\]
(d) Curves \(y = \sin(x)\) and \(y = \cos(x)\), and lines \(x = \pi/4\) and \(x = \pi/2\):
To find the limits of integration for x and y, we'll first determine the intersection points of the curves and lines:
1. \(y = \sin(x)\) and \(y = \cos(x)\) intersect when \(\sin(x) = \cos(x)\).
Solving for x, we find \(x = \pi/4\).
So, the limits of integration are as follows:
\[x: \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\]
\[y: \sin(x) \leq y \leq \cos(x)\]
Now, we can set up the double integral:
\[A = \int_{\pi/4}^{\pi/2} \int_{\sin(x)}^{\cos(x)} dy dx\]
You can now compute each of these double integrals to find the area of the respective regions.
Explanation: