Iterated Integral in Polar Coordinates

[harpazo]

Evaluate the iterated integral by converting to polar coordinates.

Let S S = interated integral symbol

S S xy dy dx

The inner integral limits are 0 to sqrt{2x - x^2}.

The outer integral limits are 0 to 2.

Solution:

I first decided to rewrite sqrt{2x - x^2} in polar form.

So, sqrt{2x - x^2} in polar form is r = 2cos(theta).

I then graphed sqrt{2x - x^2}.

From the graph, I determined that the limits of integration are follows:

r = 0 to 2cos(theta)

theta = 0 to pi/2

After working out the math, my answer is 4/5.

The textbook's answer is 2/3.

Help...

 

[MarkFL]

\(\displaystyle I=\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\int_0^{2\cos(\theta)} r^3\,dr\,d\theta\)

\(\displaystyle I=\frac{1}{4}\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\left(\left.r^4\right|_0^{2\cos(\theta)}\right)\,d\theta\)

\(\displaystyle I=4\int_0^{\frac{\pi}{2}} \sin(\theta)\cos^5(\theta)\,d\theta\)

Let:

\(\displaystyle u=\cos(\theta)\implies du=-\sin(\theta)\,d\theta\)

\(\displaystyle I=4\int_0^1 u^5\,du=\frac{4}{6}\left(\left.u^6\right|_0^1\right)=\frac{2}{3}\)

In the future, please don't begin a new thread to continue working on a problem you have previously posted. :D

 

[harpazo]

\(\displaystyle I=\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\int_0^{2\cos(\theta)} r^3\,dr\,d\theta\)

\(\displaystyle I=\frac{1}{4}\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\left(\left.r^4\right|_0^{2\cos(\theta)}\right)\,d\theta\)

\(\displaystyle I=4\int_0^{\frac{\pi}{2}} \sin(\theta)\cos^5(\theta)\,d\theta\)

Let:

\(\displaystyle u=\cos(\theta)\implies du=-\sin(\theta)\,d\theta\)

\(\displaystyle I=4\int_0^1 u^5\,du=\frac{4}{6}\left(\left.u^6\right|_0^1\right)=\frac{2}{3}\)

In the future, please don't begin a new thread to continue working on a problem you have previously posted. :D

When you did a u-sub, du = - sin(theta)d(theta).

What happened to the negative sign?

You did not include the negative sign in the integrand.

 

[MarkFL]

When you did a u-sub, du = - sin(theta)d(theta).

What happened to the negative sign?

You did not include the negative sign in the integrand.

This is what I did more fully explained...let's begin at this point:

\(\displaystyle I=4\int_0^{\frac{\pi}{2}} \sin(\theta)\cos^5(\theta)\,d\theta\)

Let:

\(\displaystyle u=\cos(\theta)\implies du=-\sin(\theta)\,d\theta\)

Okay, we may write:

\(\displaystyle I=-4\int_0^{\frac{\pi}{2}} \cos^5(\theta)(-\sin(\theta))\,d\theta\)

Now, the limits of integration are in terms of $\theta$, but we want them now in terms of $u$, so we use:

\(\displaystyle u(\theta)=\cos(\theta)\)

And so:

\(\displaystyle u(0)=\cos(0)=1\)

\(\displaystyle u\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0\)

And so after rewriting the integral in terms of $u$, we have:

\(\displaystyle I=-4\int_1^0 u^5\,du\)

Now, using the rule (which follows immediately from the anti-derivative form of the FTOC):

\(\displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\)

We now have:

\(\displaystyle I=4\int_0^1 u^5\,du\)

\(\displaystyle I=4\int_0^1 u^5\,du=\frac{4}{6}\left(\left.u^6\right|_0^1\right)=\frac{2}{3}\)

 

[harpazo]

This is what I did more fully explained...let's begin at this point:

\(\displaystyle I=4\int_0^{\frac{\pi}{2}} \sin(\theta)\cos^5(\theta)\,d\theta\)

Let:

\(\displaystyle u=\cos(\theta)\implies du=-\sin(\theta)\,d\theta\)

Okay, we may write:

\(\displaystyle I=-4\int_0^{\frac{\pi}{2}} \cos^5(\theta)(-\sin(\theta))\,d\theta\)

Now, the limits of integration are in terms of $\theta$, but we want them now in terms of $u$, so we use:

\(\displaystyle u(\theta)=\cos(\theta)\)

And so:

\(\displaystyle u(0)=\cos(0)=1\)

\(\displaystyle u\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0\)

And so after rewriting the integral in terms of $u$, we have:

\(\displaystyle I=-4\int_1^0 u^5\,du\)

Now, using the rule (which follows immediately from the anti-derivative form of the FTOC):

\(\displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\)

We now have:

\(\displaystyle I=4\int_0^1 u^5\,du\)

\(\displaystyle I=4\int_0^1 u^5\,du=\frac{4}{6}\left(\left.u^6\right|_0^1\right)=\frac{2}{3}\)

Thanks. Your reply to each question is interestingly unique and detailed. This is what I seek from every person here. I am learning calculus 3 on my own. I am not a student. Your detailed answers are wonderful and helpful.