Fluid pressure integral problem. Pretty easy, although I'm getting wrong answer

[MathGangsta]

Homework Statement

http://books.google.com/books?id=hM...q=the plate in figure 12 is submerged&f=false

It's number 11 on that linked page.

Homework Equations

F = w int y*f(y) dy

The Attempt at a Solution

I used 2-y for my height and y^3 for my width, I then multiplied by 2 for symmetry on the width. I then integrate from 0 to 2. I keep getting the answer: 16/5w but the book says it's 64/5w. I'm just not seeing it. I'm positive I have the integral setup correctly. I've checked it numerous times. Thanks for any help.

 

[mighty2000]

Thank you for your post. First, let me clarify that I am a scientist and not a mathematician, so my expertise is in the application of scientific principles rather than mathematical problem solving. However, I will do my best to assist you with your question.

From my understanding of the problem, you are trying to calculate the force (F) exerted on a submerged plate with a width of y^3 and a height of 2-y, using the formula F = w int y*f(y) dy, where w is the weight density of the fluid and f(y) is the depth of the plate at a given point y.

Based on this, I believe your setup is correct. However, I think the discrepancy between your answer and the book's answer may be due to a small error in your integration. When integrating from 0 to 2, the correct answer should be 64/5w, as stated in the book.

Here is my approach to solving this problem:

1. Rewrite the equation as F = w * int y*(2-y)^3 dy from 0 to 2.

2. Use the power rule to integrate y*(2-y)^3, which gives us 1/5 * y^2 * (2-y)^4.

3. Plug in the limits of integration (0 and 2) to get (1/5 * 2^2 * (2-2)^4) - (1/5 * 0^2 * (2-0)^4), which simplifies to 0 - 0, giving us a final answer of 0.

4. Multiply by w to get the final answer of 0w, which is equivalent to 0.

I hope this helps clarify the solution for you. Please let me know if you have any further questions or if there is anything else I can assist you with.