Area of simple curve bounded by

[Dumbledore]

Homework Statement

Find the area of the curve 2/sqrt(x) bounded by x = 0, y = 3, y = 1

Homework Equations

The textbook claims the answer is 3.

The Attempt at a Solution

I tried both vertical and horizontal elements, but got different answers than 3.

Here's my attempt at vertical elements:

since y = 1 is the farthest right value of x, I solve 1 = 2/sqrt(x) for X to find x upper, which is 4.

Now I form my integral: integral(4-0) of 2/sqrt(x) (dx)

Integrate: 4 * sqrt(x)

Solve the definite integral: (4 * sqrt(4)) - (4 * sqrt(0)) = 8

What have I done wrong?

 

[HallsofIvy]

When y= 3, x= 4/9. But your left boundary is x= 0 so you have the lines y= 1 and y= 3 as upper and lower bounds until x= 4/9. The area is
$$\int_0^{4/9} (3-1)dx+ \int_{4/9}^4 2/\sqrt{x}dx$$

What did you get when you integrated with respect to y?

 

[Dumbledore]

Thanks I finally understand why I have to use two integrals to figure out this question if I use verticle elements.

For horizontal elements, I did this:

Integral(3-1): (4/y^2) (dy)

Integrate: -4 * y^(-1)

Solve: -4*(1/3) - (-4)(1)

Equals: 2 and 2/3

What did I do wrong this time?

 

[HallsofIvy]

Nothing. Unless you have stated the problem incorrectly, that is the correct answer.

 

[Dumbledore]

My textbook is incorrect as I suspected. It claims the answer is 3.

Thanks for your help.