Finding the area between 2 curves

[ntox101]

Homework Statement

a.) Find the area of the region bounded by the graphs of f(x) = 1/x and 2x+2y=5

b.) Also, use the shell method to setup the integral that represents the volume of the solid formed by revolving the region bounded by the two same graphs about the y= 1/2. (Do not evaluate the integral)

Homework Equations

For area between two curves : Integral[f(x) - g(x)] dx (depending on which curve is on top)

For shell method : 2pi Integral[ p(x) h(x) ] dx

The Attempt at a Solution

a.) What I did was take the second equation (2x+2y=5) and solved for y to make (y = (5-2x)/2). I then made Integral[ ((5-2x)/2) - (1/x) ] dx. For [a,b], I just used the zoom function on my calculator and got a = 1/2 and b = 2. (I'm not sure if it is correct or not) After integrating, I got 1/2[5x-x²-ln|x|]

b.) I haven't been able to get this setup to the way that I think is correct. My answer is
2pi Integral[ ( (5-2y)/2 ) * (2- (1/y)) ] dy . (with a = 1/2 and b = 2). I think my [a,b] are wrong because this is in terms of y, not x, since we are revolving around the y-axis.

Also, without using Mathematica, how can I use the Integrand symbols to make this look more presentable?

Thanks,
Jon

 

[CompuChip]

For a) it looks right. By the looks of it, you are also right about the intersection points, although it would be neater if you proved that (set 1/x equal to (5 - 2x)/2 and solve for x) or at least say that you read off 1/2 and 2 and plug them into 1/x and (5 - 2x)/2 to show that they are indeed intersection points.

For b), you might find this link helpful.

 

[semc]

For part (a) i got 1/2[5x-x^2]-ln|x|
For part (b) to revolve around y-axis you need to turn the equation into x^2 = f(y)
and since you are revolving around y=1/2, you need to substitute x^2 = f(y-1/2).
Hope that helps