Question
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.
Answers
Steve
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
Try this:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F4,+y%3D4x,+y%3D8%2Fx
Steve
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
a = ∫[0,√32] 8/y - y/4 dy
Steve
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
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