# Double Integral problem

1. Sep 2, 2014

### eprparadox

1. The problem statement, all variables and given/known data
Find the volume under the surface z = y(x+2) and over the area bounded by y+x = 1, y = 1 and y = sqrt(x)

2. Relevant equations

3. The attempt at a solution

Based on the geometry of the bounds, I broke this integral into two parts. I first found the intersection of the sqrt(x) and y = 1-x. This is at x = $\dfrac {1} {2}\left( 3-\sqrt {5}\right)$

Then my first integral was this:

$$\int _{x=0}^{x=\dfrac {1} {2}\left( 3-\sqrt {5}\right) }\int _{y=1-x}^{y=1}y\left( x+2\right) dydx$$

and my second integral is this:

$$\int _{x=\dfrac {1} {2}\left( 3-\sqrt {5}\right) }^{x=1}\int _{y=\sqrt {x}}^{y=1}y\left( x+2\right) dydx$$

However when I solve this integral, I don't get the correct answer (which should be 9/8 according to the solution).

Where am I going wrong?

Thanks!

2. Sep 2, 2014

### Zondrina

By plotting the region it looks like you should be using four double integrals.

Last edited: Sep 2, 2014
3. Sep 2, 2014

### Ray Vickson

I assume you want to bound the volume below by the plane z = 0.

What was your final answer? I do not get the value 9/8, either. In fact, since 9/8 > 1 it cannot be the correct answer, for even getting a crude upper bound gives an answer < 1. Below, let
$$a = \frac{3}{2}-\frac{\sqrt{5}}{2} \doteq 0.381966012$$
be x-value of the intersection between the square root and the slanted line.

The base of the first integral is the right-triangle with sides $a$ and $a$, so has area $A_1 = a^2/2 \doteq 0.07294901715.$ A simple upper bound on $z$ is $z \leq 3$ for all feasible $x,y$, so the first integral, $I_1$, has the simple upper bound $I_1 \leq U_1 = 3 A_1 \doteq 0.2188470514$.

The base of the second integral is a "triangle" with a curved hypotenuse, so has area less than a true triangle with the same base and height; that is we have:
$$\text{area}_2 < A_2 \equiv a (1-a)/2 \doteq 0.1180339888$$
Thus, an upper bound on the second integral, $I_2$, is $I_2 \leq U_2 = 3A_2 \doteq 0.3541019664$. An upper bound on the whole double integral is $U = U_1 + U_2 \doteq 0.5729490180.$ This is a lot smaller than the claimed value of 9/8 = 1.125.

Last edited: Sep 2, 2014
4. Sep 2, 2014

### LCKurtz

Also, I would point out that, given the shape of that region, a dxdy order of integration allows setting up only one double integral.

5. Sep 2, 2014

### eprparadox

@Zondrina: I don't think there are 4 double integrals because this area is bounded by y = 1 and comes to a point at the intersection of y = 1 - x and y = sqrt(x)

@Ray Vickson: yeah I get a number much smaller than 1. I calculated the integrals on wolfram alpha and I ended up getting something like .38

@LCKurtz: I'm not sure I understand what you mean. Care to explain more?

6. Sep 3, 2014

### LCKurtz

@LCKurtz: I'm not sure I understand what you mean. Care to explain more?

I mean put the proper limits on$$\iint y(x+2)~dxdy$$Just one double integral is necessary.

7. Sep 3, 2014

### Zondrina

Yes I didn't notice this when I was sleepy. The region looks like it's divided up into three pieces, not four (somehow my brain convinced me there was a line at $x=1$ last night); The left triangle like piece, the upper triangular like piece and the bottom piece.

The limits for $x$ appear to be different in each region, as are the limits for $y$. If I'm seeing this now, it looks like you would need five integrals. Not quite sure how you would do this with just one integral.

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8. Sep 3, 2014

### Ray Vickson

Intuitively, you have chosen to split up the base area into thin vertical slices parallel to the y-axis. Essentially, the suggestion of LCKurtz is that you split it up instead into thin horizontal slices parallel to the x-axis.

9. Sep 3, 2014

### LCKurtz

@Zondrina: The only region that is described in the original post is the upper one, and it takes one double integral if done as a dxdy integral.

10. Sep 3, 2014

### Zondrina

I see all the lines only have one region where they all form a closed boundary.

Thank you for clarifying that.

Last edited: Sep 3, 2014
11. Sep 3, 2014

### eprparadox

Hey, thanks so much! Makes sense now.

Thanks!

12. Sep 3, 2014

### Ray Vickson

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