Find the area of the region bounded by the curves y = x^(-1/2), y = x^(–2), y = 1, and y = 3.
a) (1/2)(3)^1/2 + (4/3)
b) 2*(3)^1/2 - (8/3)
c) (1/2)(3)^1/2 - (32/3)
d) 2*(3)^1/2 - (32/3)
e) (8/3) - 2*(3)^1/2
So one thing that is throwing me off on this question is that I believe y = x^(-1/2) is already within the shape created by y = x^(–2) so I'm unsure whether or not I can just ignore it. I decided to do the problem without it by just solving y = x^(–2) and then using the limits of y=1 and y=3 to solve for the area. Doing this I got 4(3)^1/2)-4 which is not an answer. I'm unsure what exactly I'm doing wrong and would really appreciate some help! Thank you!
2 answers
I should probably specify that I got the area by integrating y^(-1/2) - (-y^(-1/2)) with my calculator on the interval [1,3] Sorry!
so your equations can be written as
y = 1/√x and y = 1/x^2
let's look at their graphs:
http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F%E2%88%9Ax+,+y+%3D+1%2Fx%5E2
ignore the part for x<0 , since in y = 1/√x, x > 0
so let's take horizontal slices from y = 1 to y = 3
y = 1/√x ---> y^2 = 1/x ---> x = 1/y^2
y = 1/x^2
x^2 = 1/y
x = 1/√y ----> x = y^(-1/2)
area = ∫ (y^(-1/2) - y^-2 ) dy from y = 1 to 3
= [ 2 y^(1/2) + y^-1 ] from 1 to 3
= [ 2√y + 1/y] from 1 to 3
= 2√3 + 1/3 - (2√1 + 1)
= 2√3 -8/3
y = 1/√x and y = 1/x^2
let's look at their graphs:
http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F%E2%88%9Ax+,+y+%3D+1%2Fx%5E2
ignore the part for x<0 , since in y = 1/√x, x > 0
so let's take horizontal slices from y = 1 to y = 3
y = 1/√x ---> y^2 = 1/x ---> x = 1/y^2
y = 1/x^2
x^2 = 1/y
x = 1/√y ----> x = y^(-1/2)
area = ∫ (y^(-1/2) - y^-2 ) dy from y = 1 to 3
= [ 2 y^(1/2) + y^-1 ] from 1 to 3
= [ 2√y + 1/y] from 1 to 3
= 2√3 + 1/3 - (2√1 + 1)
= 2√3 -8/3
4 answers