what is the area between the region bounded by y= x^2 , y= 162/(x^2+9) ?

Answer 1

first you can set the equations equal and find the intersections. then plug in a number for x between these two intersections (this will tell you which function is on top) then find the integral from first intersection to second intersection of top function minus bottom function. the answer= the area between the curves.

Answer 2

y = x^2, y = 162 / (x^2 + 9)

The first thing we need to do is determine our bounds for integration. What we have to do is determine where these two curves intersect, and this is done by equating them to each other.

x^2 = 162/(x^2 + 9)

Multiply both sides by x^2 + 9,

x^2(x^2 + 9) = 162
x^4 + 9x^2 = 162
x^4 + 9x^2 - 162 = 0

Therefore

(x^2 + 18) (x^2 - 9) = 0

This implies x^2 + 18 = 0 or x^2 - 9 = 0.
x^2 + 18 = 0 will have no real solutions, so we work with the other equation.
x^2 - 9 = 0, so (x - 3)(x + 3) = 0, and x = {-3, 3}

This means our bounds for integration are {-3, 3}.

Now that we have our bounds for integration, our next goal is to determine which curve is higher on the graph, from -3 to 3. The best way to do this is to test a point between -3 and 3.

Let f(x) = x^2 and g(x) = 162/(x^2 + 9). Test x = 0 (since it lies in the interval). Then
f(0) = 0^2 = 0
g(0) = 162/(0^2 + 9) = 162/9, which is a positive number.

Therefore, g(x) is clearly greater on that interval, and our formula becomes

A = Integral (-3 to 3, g(x) - f(x) ) dx

A = Integral (-3 to 3, 162/(x^2 + 9) - x^2 ) dx

We can actually split this up into two integrals.

A = Integral (-3 to 3, 162/(x^2 + 9)) dx -
Integral (-3 to 3, x^2) dx

Pulling out the constant,

A = 162 * Integral (-3 to 3, 1/(x^2 + 9) ) dx - Integral (-3 to 3,x^2 dx)

We can solve the first integral by trig substitution.
I'm not going to show you the details of this, but I can tell you right now that the integral of 1/(x^2 + 9) is going to be (1/3) arctan(x/3).

Therefore,

A = 162 arctan(x/3) {evaluated from -3 to 3} -
(1/3)x^3 {evaluated from -3 to 3}

A = 162 [arctan(1) - arctan(-1)] - (1/3) [ (3)^3 - (-3)^3 ]

A = 162 [pi/4 - (-pi/4)] - (1/3) [27 - (-27)]
A = 162 [pi/2] - (1/3) (54)
A = 81pi - 18

Answer 3

Intersection points are given by :
x^2 = 162/(x^2 + 9)
This becomes :
x^4 + 9x^2 - 162 = 0
or, (x^2 + 18)(x^2 - 9) = 0
or, (x^2 + 18)(x + 3)(x - 3) = 0
Thus, the only real intersection
points are at x = -3 and x = 3.

Turning to the graphs, we find they are
both symmetrical about the Y-axis.
Therefore, we only need to find the area
to the right of the Y-axis and double it.

The total area will be 2(A + B), where
A = [area under 162 / (x^2 + 9) from 0 to 3
minus the area of rectangle 9*3 (= 27)]
and B = [area of rectangle 9*3 (= 27)
minus the area under x^2 from 0 to 3].

Integral of 162/(x^2 + 9)
= 162(1/3)arctan(x/3)
= 54arctan(x/3)
Therefore, A = [54arctan(3/3) - 54arctan(0/3)] - 27
= 54pi/4 - 0 - 27 = 27pi/2 - 27

Integral of x^2 = x^3/3
Therefore, B = 27 - [3^3/3 - 0^3/3] = 18

Thus, 2(A + B) = 2(27pi/2 - 27 + 18) = 27pi - 18,
which is approximately 66.823.

Answer 4

The curves intersect when y= 162/(y+9) or y^2+9y -162 = 0
y = [-9 +/- sqrt( 81+4*162)]/2 = -9/2 +/- 27/2
So y = - 18 or 9 --> x = 3 or -3. So the limits of integration will be
-3 to +3
162integ(dx/(x^2+9) - integx^2 dx evaluate from -3 to +3 is area.
162(1/3tan^-1 (x/3)) - x^3/3 evaluated from -3 to +3.
162(1/3tan^-1(3/3)) - 162(1/2tan^-1(-3/3)) -[3^3/3 - (-3)^3/3)]
= 162(.262 -(-.262)) -(9 +9)
=162(.424) -18 = 66.89 units^2

Please not that the integral of dx/(x^2+9) = 1/3 arctan(x/3) and is not just arctan(x/3).

Answer 5

243 unit square