Sketch the region in the first quadrant enclosed by y=8/x, y=4x, and y=1/4x. Decide whether to integrate with respect to x or y. Then find the area of the region. Can someone please help me out with showing the work and not giving me some BS speech.
as for the area, it's best to use horizontal strips (why?) of thickness dy. Then the area is
a = ∫[0,√32] 8/y - y/4 dy
Oops. That's not quite right. Can you spot my error?
Also, it is no better to take horizontal rather than vertical strips. For vertical strips
a = ∫[0,√2] 4x - x/4 dx
+ ∫[√2,√32] 8/x - x/4 dx
To sketch the region in the first quadrant enclosed by the three curves, we need to first graph each of the curves separately.
First, let's graph the equation y = 8/x. This curve passes through the point (1, 8), where x = 1 and y = 8. As x increases, y decreases, and vice versa. Also, note that this equation is undefined when x = 0. Therefore, the graph of y = 8/x consists of a hyperbola with its branches in the first and third quadrants and asymptotes at x = 0 and y = 0.
Next, let's graph the equation y = 4x. This equation represents a straight line passing through the origin (0, 0) with a slope of 4. This line starts at the origin and extends towards the positive x and y directions.
Lastly, let's graph the equation y = 1/4x. This equation also represents a straight line passing through the origin, but with a slope of 1/4. Similar to the previous line, this line starts at the origin and extends towards the positive x and y directions.
Now, to determine whether to integrate with respect to x or y, we can analyze the intersection points of the curves.
The straight lines y = 4x and y = 1/4x intersect at (2, 8) and (4, 8). The hyperbola y = 8/x intersects with both straight lines at various points.
Let's find the intersection points between y = 8/x and y = 1/4x:
8/x = 1/4x
32 = x^2
x = ±√32
Since we are looking for the region in the first quadrant, we only consider the positive value, x = √32.
Similarly, let's find the intersection points between y = 8/x and y = 4x:
8/x = 4x
4x^2 = 8
x^2 = 2
x = ±√2
Again, considering only the positive value, x = √2.
So, the region bounded by the curves lies between x = √2 and x = √32 in the first quadrant.
To find the area enclosed, we need to integrate the appropriate function while subtracting them to get the correct area.
Since the region is bounded by y = 8/x at the top and y = 1/4x at the bottom, we will integrate with respect to x.
The area can be calculated as follows:
Area = ∫[√2, √32] (8/x - 1/4x) dx
Simplifying and integrating:
Area = ∫[√2, √32] (32 - x^2)/(4x) dx
Area = (1/4) ∫[√2, √32] (8 - x^2/x) dx
Area = (1/4) ∫[√2, √32] (8x - x^2)/x dx
Area = (1/4) ∫[√2, √32] (8 - x) dx
Now, you can proceed to solve this integral to find the exact area of the region between the curves in the first quadrant.
you are taking calculus, and you cannot draw two lines and an hyperbola? There are lots of online graphing sites; maybe you should spend some time working on this, rather than just carping and waiting.
Try this:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F4,+y%3D4x,+y%3D8%2Fx