MHB How Do You Set Up Double Integrals for Different Orders of Integration?

[harpazo]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S y/(1 + x^2) dA

R: region bounded by y = 0, y = sqrt{x}, x = 4

I can graph the region but have no idea how to proceed from there. I need solution steps.

 

[Prove It]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S y/(1 + x^2) dA

R: region bounded by y = 0, y = sqrt{x}, x = 4

I can graph the region but have no idea how to proceed from there. I need solution steps.

First we should note that the functions intersect at (0, 0), so if we use vertical strips we see that they will go from x = 0 to x = 4. The bottom of each strip is bounded by the x-axis (y = 0) and the top of each strip is bounded by the function $\displaystyle \begin{align*} y = \sqrt{x} \end{align*}$. So that means our region of integration is

$\displaystyle \begin{align*} 0 \leq y \leq \sqrt{x} \end{align*}$ with $\displaystyle \begin{align*} 0 \leq x \leq 4 \end{align*}$

 

[topsquark]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S y/(1 + x^2) dA

R: region bounded by y = 0, y = sqrt{x}, x = 4

I can graph the region but have no idea how to proceed from there. I need solution steps.

Have you considered that if you post these one at a time you might be able to solve some on your own? You have posted five threads with pretty much the exact same question.

-Dan

 

[harpazo]

Thank you everyone.

 

[MarkFL]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S y/(1 + x^2) dA

R: region bounded by y = 0, y = sqrt{x}, x = 4

I can graph the region but have no idea how to proceed from there. I need solution steps.

Let's first look at the region $D$ over which we are directed to integrate:

View attachment 6551

Now if we use horizontal strips, then we see they are bound on the left by $x=y^2$ and on the right by $x=4$. We also see that these horizontal strips run from $y=0 $to $y=2$. And so we may write the integral as:

$$I=\int_0^2 y\int_{y^2}^4 \frac{1}{x^2+1}\,dx\,dy$$

Evaluating (I know you asked not to evaluate, but I figure enough time has gone by, and I wish to check to be sure we get the same result regardless of integration order), we obtain:

$$I=\int_0^2 y\arctan(4)-y\arctan\left(y^2\right)\,dy=\frac{1}{4}\ln(17)$$

Now, if we set this up with vertical strips, we see that they are bound on the bottom by $y=0$ and on the top by $y=\sqrt{x}$, and that these strips run from $x=0$ to $x=4$. And so we may write the integral as:

$$I=\int_0^4 \frac{1}{x^2+1}\int_0^{\sqrt{x}} y\,dy\,dx$$

Evaluating, we obtain:

$$I=\frac{1}{4}\int_0^4\frac{2x}{x^2+1}\,dx=\frac{1}{4}\ln(17)$$

I found this method simpler to integrate. :D

 

[harpazo]

Let's first look at the region $D$ over which we are directed to integrate:
Now if we use horizontal strips, then we see they are bound on the left by $x=y^2$ and on the right by $x=4$. We also see that these horizontal strips run from $y=0 $to $y=2$. And so we may write the integral as:

$$I=\int_0^2 y\int_{y^2}^4 \frac{1}{x^2+1}\,dx\,dy$$

Evaluating (I know you asked not to evaluate, but I figure enough time has gone by, and I wish to check to be sure we get the same result regardless of integration order), we obtain:

$$I=\int_0^2 y\arctan(4)-y\arctan\left(y^2\right)\,dy=\frac{1}{4}\ln(17)$$

Now, if we set this up with vertical strips, we see that they are bound on the bottom by $y=0$ and on the top by $y=\sqrt{x}$, and that these strips run from $x=0$ to $x=4$. And so we may write the integral as:

$$I=\int_0^4 \frac{1}{x^2+1}\int_0^{\sqrt{x}} y\,dy\,dx$$

Evaluating, we obtain:

$$I=\frac{1}{4}\int_0^4\frac{2x}{x^2+1}\,dx=\frac{1}{4}\ln(17)$$

I found this method simpler to integrate. :D

Wonderfully done!