Multivariable calculus -double int

[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?

 

[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

 

[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?

 

[HallsofIvy]

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?

 

[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

 

[HallsofIvy]

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?"

 

[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}
Answer is 52 units cubed.

 

[HallsofIvy]

\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
which doesn't seem to give what you have.