Double integral -II

[canvas]

Yeah it's clear now, if we take x or y greater than 1 then sqrt(x) or sqrt(y) will be greater than 1, and since sqrt(x) and sqrt(y) are greater or equal than 0 so the sum of sqrt(x)+sqrt(y) will be greater than 1, but it cannot be, right?

 

[BigBeachBanana]

if we take x or y greater than 1 then sqrt(x) or sqrt(y) will be greater than 1,

Agreed.

since sqrt(x) and sqrt(y) are greater or equal than 0 so the sum of sqrt(x)+sqrt(y) will be greater than 1, but it cannot be, right?

We established that x must be between 0 and 1. So for whatever x-value we pick, y must be "adjusted" to meet the inequality <=1. So it will never exceed 1. For example, x=0 then 0<=y<=1 OR if x=1 then y=0.

 

[canvas]

Maybe I wasn't clear so I'll post my thoughts in latex here:

 

[nasi112]

"X's limits only range from 0-1."

Can't see that ;/

You can look at the nice graph that Beach uploaded, or you can test some values for $\displaystyle x$.

$\displaystyle \sqrt{y} \leq 1- \sqrt{x}$

when $\displaystyle x = 0$
$\displaystyle \sqrt{y} \leq 1$, we are fine.

when when $\displaystyle x = 1$
$\displaystyle \sqrt{y} \leq 0$, we are fine.

when when $\displaystyle x = 2$
$\displaystyle \sqrt{y} \leq 1 - \sqrt{2} \ = \$negative, we are not Fine.

So maximum number for $\displaystyle x$ is $\displaystyle 1$.

 

[canvas]

You can look at the nice graph that Beach uploaded, or you can test some values for $\displaystyle x$.

$\displaystyle \sqrt{y} \leq 1- \sqrt{x}$

when $\displaystyle x = 0$
$\displaystyle \sqrt{y} \leq 1$, we are fine.

when when $\displaystyle x = 1$
$\displaystyle \sqrt{y} \leq 0$, we are fine.

when when $\displaystyle x = 2$
$\displaystyle \sqrt{y} \leq 1 - \sqrt{2} \ = \$negative, we are not Fine.

So maximum number for $\displaystyle x$ is $\displaystyle 1$.

look at #23 there is my explanation

 

[nasi112]

look at #23 there is my explanation

Yeah, nice explanation

I think that you are now good at inequalities as well as double integrals.

Inequality is a tricky subject, so whenever you have questions about it, try to graph or test some values.

For example, $\displaystyle x^2 + x - 2 \leq 0$. Imagine that you are in a test and you cannot graph this function. What are the range of $\displaystyle x$ values?

$\displaystyle ? \leq x \leq \ ?$

Think about this question, and if you could answer it without seeing the graph, you are ahead of tons of students.

 

[canvas]

We don't have to graph it it's a quadratic inequality, we know how to solve it easily:

x²+x-2 = x²+2x-x-2 = x(x+2)-(x+2) = (x+2)(x-1)
So (x+2)(x-1) ≤ 0 <=> x ∈ [-2,1]

 

[canvas]

Or I didn't got ur question?

 

[BigBeachBanana]

So what answer did you get?

 

[canvas]

So what answer did you get?

Here is my attempt, thank you Sir for all the time you spent explaining me these things

 

[nasi112]

Here is my attempt, thank you Sir for all the time you spent explaining me these things

View attachment 30791

wOW

We don't have to graph it it's a quadratic inequality, we know how to solve it easily:

x²+x-2 = x²+2x-x-2 = x(x+2)-(x+2) = (x+2)(x-1)
So (x+2)(x-1) ≤ 0 <=> x ∈ [-2,1]

Correct?