# Multivariable calculus -double int

1. Jul 29, 2006

### ultima9999

R is the interior of the region, in the x,y plane, bounded by the parabola $$y = 4 - (x - 3)^2$$ and for which $$x \leq 3\,\mbox{and}\,y \geq 0$$.
Sketch the region R, and evaluate the double integral $$\iint_R 2xy\,dx\,dy$$

I've drawn the region, but I am unsure as to what to do with the integral and how R links into it. Do I simply make 3 and 0 the upper and lower limits for x, and make 4 and 0 the upper and lower limits for y when integrating? ie. $$\int_{0}^{4}\int_{0}^{3} 2xy\,dx\,dy$$?

Last edited: Jul 29, 2006
2. Jul 29, 2006

### benorin

No, the domain of integration of $$\int_{0}^{4}\int_{0}^{3} 2xy\,dx\,dy$$ is a rectangle, having (0,0) and (3,4) as opposite corners. The given region R is parabolic, and easily described in the form $$f_1(y) \leq x\leq f_2(y), a\leq y\leq b$$. To determine the two functions bounding x, look at the picture of R: for each fixed y value, x ranges between what curve (as function of y) and 3? after that, the bounds on y are constants: $$0\leq y\leq 4$$

3. Jul 30, 2006

### ultima9999

I still don't really understand how I should start off the question. x is bounded by $$y = 4 - (x - 3)^2\,\mbox{and}\,y = 3$$. So my equation would be $$\int_{0}^{4}\int_{\sqrt{4 - y}}^{3} 2xy\,dx\,dy$$?

4. Jul 30, 2006

### HallsofIvy

Staff Emeritus
You said you'd drawn the graph- that should tell you all you need to know. You should see that the parabola is symmetric about x= 3 and crosses the x-axis at x= 1 and x= 5. Those are important because you are requiring that $y\ge 0$. Also, because you are requiring that $x\le 3$, the region over which you are integrating is the left half of that parabola- only up to the axis of symmetry.

The way you have it set up, you are doing the "inner" integral with respect to x and the "outer" integral with respect to y. That's perfectly fine. Since the limits on the "outer" integral must be constants, what are the smallest and largest possible values of y in that region? Yes, they are 0 and 4, exactly as you have. For each y, then, what are the limits on x? Imagine drawing a horizontal line across the parabola. The lower value for x is on the parabola while the higher value is on the axis of symmetry, x= 3. Clearly the upper limit of integration is 3, as you have. What about the lower limit? That's on the parabola given by y= 4- (x-3)2. Solve that for x (it is not $\sqrt{4- x}$!). That quadratic equation will have two roots and you will have to decide which to choose.

As I said, that choice of order of integration is perfectly valid but I suspect most people would have chosen the other way: integrate with respect to y first and then x. What is the smallest value x takes on in that region? What is the largest? Now draw a vertical line at some representative value of x. What are the lowest and highest values of y in terms of x?

5. Jul 30, 2006

### ultima9999

$$x = \sqrt{4 - y} + 3$$
$$x = \pm2 + 3 = 1\,\mbox{or}\,5$$
As $$x \leq 3$$ choose x = 1.

After solving the integral, I get:
$$\int_{1}^{3}\int_{0}^{4} 2xy\,dy\,dx = 64$$

6. Jul 30, 2006

### HallsofIvy

Staff Emeritus
In spite of the fact that you were just told that was wrong??
Integrating x from 1 to 3 and y from 0 to 4 integrates over the rectangle bounded by x=1, x=3, y= 0, y= 4. You were told that in benorin's first response. Unless the region of integration is a rectangle, the inner integral must depend on the second variable.
To repeat (this time with emphasis!)
"Now draw a vertical line at some representative value of x. What are the lowest and highest values of y in terms of x?"

7. Jul 31, 2006

### ultima9999

Ok, I think I got you. I chose to integrate with respect to x first.

$$\int_{0}^{4}\int_{-\sqrt{4-y}+3}^{3} 2xy\,dx\,dy$$
$$\int_{0}^{4}[x^2 y]_{-\sqrt{4-y}+3}^{3}\,dy$$
$$[5y^2 - 7y - 4(4 - y)^\frac{3}{2}]_{0}^{4}$$
$$\int_{0}^{4}[x^2 y]_{-\sqrt{4-y}+3}^{3}\,dy$$
$$\int_0^49y- y(3- \sqrt{4-y})^2 dy= \int_0^4y^2-4y- 6y\sqrt{4-y}dy$$