Consider the region enclosed by the graphs of x=y^2-5 and x=3-y^2

a)Express the area of this region by setting up an integral with respect to x

b) Express the area of this region by setting up an integral with respect to y

c) Find the area of this region by evaluating one of the definite integrals found above

The curves intersect at (-1,2) and (-1,-2)

So, using horizontal strips of width dy, the area is

a = ∫[-2,2] (3-y^2)-(y^2-5) dy

Using vertical strips, we have to split the region in two at the intersections, and then symmetry helps:

a = ∫[-5,-1] 2√(x+5)dx
+ ∫[-1,3] 2√(3-x)dx

To find the area of the region enclosed by the graphs of the given equations, we can set up an integral either with respect to x or with respect to y.

a) To set up an integral with respect to x, we need to express y in terms of x and determine the limits of integration.

From the equation x = y^2 - 5, we can rearrange it to express y in terms of x: y = ±√(x + 5).

Now we need to find the limits of integration by determining the x-values where the graphs intersect.

Setting the two equations equal, y^2 - 5 = 3 - y^2, we get:
2y^2 = 8
y^2 = 4
y = ±2 (two distinct y-values)

Since the graphs intersect at y = ±2, we know that the limits of integration with respect to x are x = (-5, 3).

Therefore, we can set up the integral with respect to x as:
∫[(-5, 3)] (f(x) - g(x)) dx, where f(x) is the upper function and g(x) is the lower function.

In this case, f(x) = √(x + 5) and g(x) = -√(x + 5).

So the integral for the area is:
∫[(-5, 3)] (√(x + 5) - (-√(x + 5))) dx.

b) To set up an integral with respect to y, we need to express x in terms of y and determine the limits of integration.

From the equation x = 3 - y^2, we can rearrange it to express x in terms of y: x = 3 - y^2.

Now we need to find the limits of integration by determining the y-values where the graphs intersect.

Setting the two equations equal, y^2 - 5 = 3 - y^2, we get:
2y^2 = 8
y^2 = 4
y = ±2 (two distinct y-values)

Since the graphs intersect at y = ±2, we know that the limits of integration with respect to y are y = (-2, 2).

Therefore, we can set up the integral with respect to y as:
∫[(-2, 2)] (f(y) - g(y)) dy, where f(y) is the right function and g(y) is the left function.

In this case, f(y) = 3 - y^2 and g(y) = y^2 - 5.

So the integral for the area is:
∫[(-2, 2)] (3 - y^2 - (y^2 - 5)) dy.

c) To find the area of the region, we can evaluate either of the definite integrals obtained from setting up the integral in parts a) or b).

Let's evaluate the definite integral obtained from part a). The integral is:
∫[(-5, 3)] (√(x + 5) - (-√(x + 5))) dx.

Evaluating this integral gives us the area of the region enclosed by the graphs.