**The area between any of the two of lines, circles/parabolas/ellipses (the region should be clearly identifiable)**

**Find the area enclosed by two sections: a line at y = x – 1 and the parabola y^{2 }= 2x + 6.**

By solving the two equations, we find that the points of intersection are (-1, -2) and (5, 4).

We use the equation of the parabola and solve for *x.*

From the figure as seen below, we deduce that the left and right boundary curves are:

We must integrate between the appropriate *y*-values as seen from the graph above, *y *=-2 and *y *=4.

To find the area between the curves *y *=* f*(*x*) and *y = g*(*x*), where *f*(*x*) ≥ *g*(*x*) for some values of *x* but *g*(*x*) ≥ *f*(*x*) for other values of *x*. We then split the given region *S* into several regions *S*_{1}, *S*_{2}, . . . with areas *A*_{1}, *A*_{2}, . . .Then, we define the area of the region *S* to be the sum of the areas of the smaller regions *S*_{1}, *S*_{2}, . . . , that is, *A *= *A*_{1 }+ *A*_{2 }+. . .

The area between the curves *y *=* f*(*x*) and *y *=* g*(*x*) and between the closed interval of *x*=* a *and *x *=* b* is:

**Example: Find the area of the region bounded by the curves y = sin x, y = cos x, x =0, and x = π/2.**

The points of intersection occur when sin x = cos x, that is, when x = π / 4 (since 0 ≤ x ≤ π / 2).

Observe that cos *x* ≥ sin *x* when 0 ≤ *x* ≤ *π*/ 4 but sin *x* ≥ cos *x* when *π*/ 4 ≤ *x* ≤ *π*/ 2.

In general:

**Find the area between an ellipse and the coordinates x = 0 and the ordinates x = 0**

and x = ae, also b2 = a2 (1 – e2) and e < 1.

The required area of the region BOB′RFSB is enclosed by the ellipse and the lines *x *= 0 and *x *= *ae*. Note that the area of the region BOB′RFSB

**Example: Find the area of the region in the 1 ^{st} quadrant between by the line y = x, a circle x^{2}+ y^{2 }= 32 and the x-axis.**

There are two equations: y = x and x^{2}+ y^{2 }= 32

Solving (1) and (2), we find that the lineand the circle meet at B(4, 4) in the 1^{st}quadrant. Draw perpendicularBM to the x-axis. Therefore, the required area = area of the region OBMO + area of the region BMAB.

Now, the area of the region OBMO

The area of the region BMAB:

= 8π – (8 + 4π) = 4π -8

Adding the two sections we get: 4π