MHB Double Integrals in Polar Coordinates

[harpazo]

Evaluate the iterated integral by converting to polar coordinates.

Let S S = double integral symbol

S S y dx dy

The outer integral is from 0 to a.

The inner integral is from 0 to sqrt{a^2 - y^2}.

I started by letting y = r sin ϴ

S S r sinϴ dxdy.

I then let dxdy = r dr d ϴ

S S r sin ϴ rdr dϴ

S S r^2 sin ϴ dr dϴ

Is this correct so far?

If not, can someone set it up for me?

The answer is (a^3)/3.

What correct set up leads to the answer given here?

 

[MarkFL]

What we have is:

$$I=\int_0^a y\int_0^{\sqrt{a^2-y^2}}\,dx\,dy$$

Evaluating the inner integral, we obtain:

$$I=\int_0^a y\sqrt{a^2-y^2}\,dy=-\frac{1}{2}\int_0^a \sqrt{a^2-y^2}(-2y)\,dy$$

Continue with the integration (no substitution needed) to get the result you posted. :)

 

[harpazo]

What we have is:

$$I=\int_0^a y\int_0^{\sqrt{a^2-y^2}}\,dx\,dy$$

Evaluating the inner integral, we obtain:

$$I=\int_0^a y\sqrt{a^2-y^2}\,dy=-\frac{1}{2}\int_0^a \sqrt{a^2-y^2}(-2y)\,dy$$

Continue with the integration (no substitution needed) to get the result you posted. :)

I can integrate this double integral without using polar coordinates but how is it done using polar coordinates?

 

[MarkFL]

I can integrate this double integral without using polar coordinates but how is it done using polar coordinates?

Okay, our first step is to transform the iterated integral into a double integral:

$$I=\int_0^a y\int_0^{\sqrt{a^2-y^2}}\,dx\,dy=\iint\limits_{R}y\,dA$$

where $R$ is a sector of a circle with radius $a$. In polar coordinates, $R$ is the region between $r=0$ and $r=a$ for $$\theta\in\left[0,\frac{\pi}{2}\right]$$.

Since $y=r\sin(\theta)$, the double integral thus becomes:

$$I=\int_{0}^{\frac{\pi}{2}} \sin(\theta) \int_0^{a} r^2\,dr\,d\theta$$

Can you proceed?

 

[harpazo]

Okay, our first step is to transform the iterated integral into a double integral:

$$I=\int_0^a y\int_0^{\sqrt{a^2-y^2}}\,dx\,dy=\iint\limits_{R}y\,dA$$

where $R$ is a sector of a circle with radius $a$. In polar coordinates, $R$ is the region between $r=0$ and $r=a$ for $$\theta\in\left[0,\frac{\pi}{2}\right]$$.

Since $y=r\sin(\theta)$, the double integral thus becomes:

$$I=\int_{0}^{\frac{\pi}{2}} \sin(\theta) \int_0^{a} r^2\,dr\,d\theta$$

Can you proceed?

Yes, I can. Thanks.

 

[HallsofIvy]

Notice that [math]x= \sqrt{4^2- y^2}[/math] is the right semi-circle of the circle [math]x^2+ y^2= 4[/math], the circle with center at (0, 0) and radius 2. Restricting y to lie between 0 and a means that we have the upper right quadrant of that circle. In polar coordinates, r will go from 0 to a while \theta goes from 0 to \pi/2. Since, as you say, in polar coordinates y= r sin(\theta) and dxdy= r dr d\theta, the integral becomes
[math]\int_0^a \int_0^{\pi/2} r^2 sin(\theta)d\theta dr= \left(\int_0^a r^2 dr\right)\left(\int_0^{\pi/2} sin(\theta)d\theta[/math]

 

[harpazo]

Notice that [math]x= \sqrt{4^2- y^2}[/math] is the right semi-circle of the circle [math]x^2+ y^2= 4[/math], the circle with center at (0, 0) and radius 2. Restricting y to lie between 0 and a means that we have the upper right quadrant of that circle. In polar coordinates, r will go from 0 to a while \theta goes from 0 to \pi/2. Since, as you say, in polar coordinates y= r sin(\theta) and dxdy= r dr d\theta, the integral becomes
[math]\int_0^a \int_0^{\pi/2} r^2 sin(\theta)d\theta dr= \left(\int_0^a r^2 dr\right)\left(\int_0^{\pi/2} sin(\theta)d\theta[/math]

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