1. Jan 30, 2015

### Math10

1. The problem statement, all variables and given/known data
Evaluate the double integral 4x^3 dA, where R is the region bounded by the graphs of y=(x-1)^2 and y= -x+3.

2. Relevant equations
None.

3. The attempt at a solution
The answer in the book is 72/5. But I got 45 as the answer.
Here's my work:
(x-1)^2= -x+3
x^2-2x+1= -x+3
x=2, -1
y=1, 4.
double integral of 4x^3 dx dy from -1 to 2 for dx and from 1 to 4 for dy
integral of x^4 (evaluate from -1 to 2) dy from 1 to 4 for dy
=45

2. Jan 30, 2015

### Staff: Mentor

No. The y values run from the parabola up to the line. They do not run from 1 to 4.

3. Jan 30, 2015

### Math10

So how do I find the limits of integration for y?

4. Jan 30, 2015

### Staff: Mentor

You get the limits from these equations. Did you draw a picture of this region?

5. Jan 30, 2015

### Math10

Since it's a double integral, the x is from -1 to 2 for dx. What about y?

6. Jan 30, 2015

### O_o

7. Jan 31, 2015

### Math10

I still don't get it. What are the limits of integration for y?

8. Jan 31, 2015

### O_o

When x is between -1 and 2 one of the functions is lower than the other. When you plot them can you see which it is? That is the lower bound. And the upper bound is the function that is higher.

The upper and lower bound varies with x. That's why you can't just use constants...you need to write the bounds in terms of functions. Try watching these 2 videos:

Last edited: Jan 31, 2015
9. Jan 31, 2015

### Staff: Mentor

I asked before, but you didn't respond. Did you sketch a graph of the region over which you're integrating?

10. Jan 31, 2015

### Math10

You know what? I got it.