# Homework Help: Volume between given planes

1. Mar 5, 2009

### alalall2

1. The problem statement, all variables and given/known data

(1 pt) Find the volume of the region bounded by the planes

2. Relevant equations
V = ∫∫7/4-6/4y-2/4x

3. The attempt at a solution
Since y=x I found their values when z = 0.
6x+2x=7, x=7/8
y= 7/8 is the maximum value y will have in this integration as it decreases as x approaches its maximum value: 2x = 7, x=7/2.

When I made a 2-D image of the region I would be integrating on I came up with a triangle which had two areas that needed to be integrated separately as in the first region: 0≤x≤7/8 and 0≤y≤x, and in the second region: 7/8≤x≤7/2 and 0≤y≤(7/6 - x/3).

After integrating my result was 0.893229167.

What I am wondering is if I am using the correct limits for my integration or if I made a mistake in my math.

Last edited by a moderator: Apr 24, 2017
2. Mar 5, 2009

### wywong

I would avoid double integral thus: a horizontal cross-section of the region is bounded by the x and y axes and the line

$$2x+6y = 7-4z$$

The area of that right angle triangle is $$\frac{(7-4z)^2}{24}$$ for z in$$[0,\frac{7}{4}]$$.

The definite integral should be $$\frac{7^3}{12\cdot 24}=1.19$$

3. Mar 5, 2009

### alalall2

The system still thinks that is an incorrect answer

4. Mar 6, 2009

### wywong

Could the system be expecting a different format than what you entered, such as different number of decimal places?