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From: Fred Nurk on 18 Jul 2010 00:22
Find the exact area of the region bounded by the graphs with equations y
^ 2 = x and x - y = 2.

int{[x^(1/2)-x+2], x = 0...4} - int{[x-2 - x^(1 / 2)], x = 0...2} =
16 / 3 - 8 + 8 - (0 - 0 + 0) - [2 - 4 - 4 * 2 ^ (1 / 2) / 3 - 0] =
[16 + 2 + 4 * 2 ^ (1 / 2)] / 3 = [18 + 4 * 2 ^ (1 / 2)] / 3.

Exact area of [18 + 4 * 2 ^ (1 / 2)] / 3 square units.

Where have I gone wrong?
From: Greg Neill on 18 Jul 2010 00:49
Fred Nurk wrote:
> Find the exact area of the region bounded by the graphs with equations y
> ^ 2 = x and x - y = 2.
>
> int{[x^(1/2)-x+2], x = 0...4} - int{[x-2 - x^(1 / 2)], x = 0...2} =
> 16 / 3 - 8 + 8 - (0 - 0 + 0) - [2 - 4 - 4 * 2 ^ (1 / 2) / 3 - 0] =
> [16 + 2 + 4 * 2 ^ (1 / 2)] / 3 = [18 + 4 * 2 ^ (1 / 2)] / 3.
>
> Exact area of [18 + 4 * 2 ^ (1 / 2)] / 3 square units.
>
> Where have I gone wrong?

Try plotting both functions on a single set of axes and
then check to see if your choice of integrands and
end points make sense.

As a further hint, try plotting the curves as functions of
x in terms of y rather than the usual y against x. That
is, plot x(y) rather than y(x).


From: William Elliot on 18 Jul 2010 00:55
On Sun, 18 Jul 2010, Fred Nurk wrote:

> Find the exact area of the region bounded by the graphs with equations y
> ^ 2 = x and x - y = 2.
>
y = sqr x; y = x - 2;
x = x^2 - 4x + 4; x^2 - 5x + 4 = (x - 1)(x - 4) = 0

> int{[x^(1/2)-x+2], x = 0...4} - int{[x-2 - x^(1 / 2)], x = 0...2} =
> 16 / 3 - 8 + 8 - (0 - 0 + 0) - [2 - 4 - 4 * 2 ^ (1 / 2) / 3 - 0] =
> [16 + 2 + 4 * 2 ^ (1 / 2)] / 3 = [18 + 4 * 2 ^ (1 / 2)] / 3.
> Exact area of [18 + 4 * 2 ^ (1 / 2)] / 3 square units.
>
> Where have I gone wrong?

Area = integral(0,4) (sqr x - (x - 2)) dx

From: Greg Neill on 18 Jul 2010 01:11
William Elliot wrote:
> On Sun, 18 Jul 2010, Fred Nurk wrote:
>
>> Find the exact area of the region bounded by the graphs with equations y
>> ^ 2 = x and x - y = 2.
>>
> y = sqr x; y = x - 2;
> x = x^2 - 4x + 4; x^2 - 5x + 4 = (x - 1)(x - 4) = 0
>
>> int{[x^(1/2)-x+2], x = 0...4} - int{[x-2 - x^(1 / 2)], x = 0...2} =
>> 16 / 3 - 8 + 8 - (0 - 0 + 0) - [2 - 4 - 4 * 2 ^ (1 / 2) / 3 - 0] =
>> [16 + 2 + 4 * 2 ^ (1 / 2)] / 3 = [18 + 4 * 2 ^ (1 / 2)] / 3.
>> Exact area of [18 + 4 * 2 ^ (1 / 2)] / 3 square units.
>>
>> Where have I gone wrong?
>
> Area = integral(0,4) (sqr x - (x - 2)) dx

y^2 = x is not quite the same thing as x = sqrt(y),
given how the domain of the square root function is
defined for real functions.

y^2 = x represents a parabola "lying on its side",
and the square root will only reproduce the portion
above the x-axis. The bottom half would be give by
the negative of the square root of x.

If you want to integrate the area using x as the
variable of integration, you'll have to be aware
that for a portion of the domain the bounds will
be the upper and lower halves of the parabola.

A sketch of the two equations should make this clear.

It might be simpler to write x in terms of y for
both equations and integrate over the appropriate
y range. That way only a single integral is
involved.


From: William Elliot on 18 Jul 2010 01:41
On Sun, 18 Jul 2010, Greg Neill wrote:
> William Elliot wrote:
>>
>>> Find the exact area of the region bounded by the graphs with equations
>>> y ^ 2 = x and x - y = 2.
>>>
>> y = sqr x; y = x - 2;
>> x = x^2 - 4x + 4; x^2 - 5x + 4 = (x - 1)(x - 4) = 0
>>
>> Area = integral(0,4) (sqr x - (x - 2)) dx
>
> y^2 = x represents a parabola "lying on its side",
> and the square root will only reproduce the portion
> above the x-axis.

x = y^2; x = y + 2
y^2 = y + 2; y^2 - y - 2 = (y - 2)(y + 1) = 0

Area = integral(-1,2) (y^2 - y - 2) dy

> It might be simpler to write x in terms of y for
> both equations and integrate over the appropriate
> y range. That way only a single integral is
> involved.
>
Indeed.
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