How Do You Set Up Double Integrals Over a Semicircular Region?

[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 (x^2 + y^2) dA

R: semicircle bounded by y = {4 - x^2}, y = 0

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 (x^2 + y^2) dA

R: semicircle bounded by y = {4 - x^2}, y = 0

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

We should first note that the two boundary functions intersect at $\displaystyle \begin{align*} (2, 0) \end{align*}$ and $\displaystyle \begin{align*} (-2, 0) \end{align*}$, so if we use vertical strips, then they will go from x = -2 to x = 2. Each strip is bounded below by y = 0 and bounded above by $\displaystyle \begin{align*} y = \sqrt{4 - x^2} \end{align*}$, so that means our region of integration is

$\displaystyle \begin{align*} 0 \leq y \leq \sqrt{4 - x^2} \end{align*}$ with $\displaystyle \begin{align*} -2 \leq x \leq 2 \end{align*}$

 

[harpazo]

Thank you everyone.

 

[MarkFL]

Let's look at the region over which we are to integrate:

View attachment 6553

If we use horizontal strips, we see they are bounded on the left by $x=-\sqrt{4-y^2}$ and on the right by $x=\sqrt{4-y^2}$, however, we find the integrand is an even function so we may use the first quadrant area, and double it. We also see the strips run from $y=0$ to $y=2$, and so our integral becomes:

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

Evaluating, we obtain:

$$I=\frac{2}{3}\int_0^2\left(\left[x^3+3xy^2\right]_0^{\sqrt{4-y^2}}\right)\,dy=\frac{2}{3}\int_0^2 (4-y^2)^{\frac{3}{2}}+3y^2(4-y^2)^{\frac{1}{2}}\,dy=4\pi$$

Now if we use vertical strips, we see they are bounded on the bottom by $y=0$ and on the top by $y=\sqrt{4-x^2}$, and run from $x=-2$ to $x=2$, although we may employ the same even function symmetry we used before, to write the integral as:

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

We see that in this integral because of the symmetry of the region over quadrants I and II, and the even-symmetry of the integrand, the two dummy variable of integration can simply be interchanged. :D

 

[harpazo]

Thank you everyone.

 

[HallsofIvy]

In the original post we are told that the region is "the semicircle bounded by y = {4 - x^2}, y = 0". But that is not a semi-circle, it is a parabolic region. Every one assumed that what was meant was [math]y= \sqrt{4- x^2}[/math]. Is that correct?

 

[harpazo]

In the original post we are told that the region is "the semicircle bounded by y = {4 - x^2}, y = 0". But that is not a semi-circle, it is a parabolic region. Every one assumed that what was meant was [math]y= \sqrt{4- x^2}[/math]. Is that correct?

Yes, you are right.
It should be y = sqrt{4 - x^2}.

 

[MarkFL]

Yes, you are right.
It should be y = sqrt{4 - x^2}.

I didn't notice before, but you share some striking similarities with another user here:

• Sometimes writes a square root as {x}.
• States that $\LaTeX$ sometimes hides behind text.
• Says "Thank you everyone."
• Has the same IP address.

Coincidence? (Thinking)

 

[harpazo]

I didn't notice before, but you share some striking similarities with another user here:

• Sometimes writes a square root as {x}.
• States that $\LaTeX$ sometimes hides behind text.
• Says "Thank you everyone."
• Has the same IP address.

Coincidence? (Thinking)

Yes, coincidence.

 

[MarkFL]

Yes, coincidence.

(Rock) (Bandit)

 

[harpazo]

Moving on...