1. ## Integrals and Centroids

Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here.

Centroids:
Find the center of mass of a semi-circular plate of radius r
Find the volume when the plate (above) is rotated around a line along its straight side

Integrals:

$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}dx$

$\displaystyle \int \frac{e^{2t}}{1+e^{4t}}dt$

$\displaystyle \int \frac{dx}{\sqrt{x^2 - 16}}$

2. Originally Posted by Debelius
Integrals:

$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}dx$

$\displaystyle \int \frac{e^{2t}}{1+e^{4t}}dt$

$\displaystyle \int \frac{dx}{\sqrt{x^2 - 16}}$

$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}dx$

Use integration by parts:

$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}\, dx = \ln x (2\sqrt{x}) - \int \frac{2\sqrt{x}}{x}\, dx = 2\sqrt{x} \ln x - 2\int \frac{1}{\sqrt{x}}\, dx = 2\sqrt{x} \ln x - 4\sqrt{x}+C$

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$\displaystyle \int \frac{e^{2t}}{1+e^{4t}}dt$

Use substitution $\displaystyle u = e^{2t}$,

$\displaystyle \int \frac{e^{2t}}{1+e^{4t}}dt = \frac12\int \frac{1}{1+u^2}du = \frac12\tan^{-1} u + C = \frac12\tan^{-1} e^{2t} + C$

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$\displaystyle \int \frac{dx}{\sqrt{x^2 - 16}}$

Use substitution $\displaystyle x = 4\sec u \Rightarrow dx = 4\sec u\tan u du$

$\displaystyle \int \frac{dx}{\sqrt{x^2 - 16}} = \int \frac{4\sec u\tan u}{4\tan u}\, du $$\displaystyle =\int \sec u \, du = \ln |\sec u + \tan u| + C = \ln \left|\dfrac{x}{4} + \frac{\sqrt{x^2 - 16}}{4}\right| + C 3. Originally Posted by Debelius Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here. Centroids: Find the center of mass of a semi-circular plate of radius r Chris says: To find the center of gravity [mass], we need to know its density function \displaystyle \delta(x,y). If you're looking for the centroid, just use the formulas \displaystyle \bar x = \frac{{\iint\limits_R {x\,dA}}} {{\iint\limits_R {dA}}} and \displaystyle \bar y = \frac{{\iint\limits_R {y\,dA}}} {{\iint\limits_R {dA}}} Find the volume when the plate (above) is rotated around a line along its straight side Chris says: We can use the Theorem of Pappus to find the volume. First find the area of the region we're dealing with, and then multiply it by the distance traveled by the centroid [in this case \displaystyle 2\pi\bar y] Hope that this makes sense! --Chris EDIT : It should \displaystyle 2\pi {\color{red}\bar y}, NOT \displaystyle 2\pi {\color{red} \bar x} as the distance traveled by the centroid (in the second question) 4. Originally Posted by Debelius Heh, I got more integrals I need help with, and the still unanswered centroid question, which I'll just include here. Centroids: Find the center of mass of a semi-circular plate of radius r Find the volume when the plate (above) is rotated around a line along its straight side Integrals: \displaystyle \int \frac{\ln(x)}{\sqrt{x}}dx \displaystyle \int \frac{e^{2t}}{1+e^{4t}}dt \displaystyle \int \frac{dx}{\sqrt{x^2 - 16}} Ok for the first one here is what I would do \displaystyle \int\frac{\ln(x)}{\sqrt{x}}~dx=2\int\frac{\frac{1} {2}\ln(x)}{\sqrt{x}}$$\displaystyle =2\int\frac{\ln\left(\sqrt{x}\right)}{\sqrt{x}}~ds =4\int\frac{\ln\left(\sqrt{x}\right)}{2\sqrt{x}}=4 \bigg[\sqrt{x}\ln(\sqrt{x})-\sqrt{x}\bigg]$

That last part was by letting $\displaystyle u=\sqrt{x}$
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For the second one rewrite it as

$\displaystyle \int\frac{e^{2x}}{1+\left(e^{2x}\right)^2}$

Now let $\displaystyle u=e^{2x}$

So $\displaystyle du=2e^{2x}$

So we have

$\displaystyle \frac{1}{2}\int\frac{du}{1+u^2}=\frac{1}{2}\arctan (u)$

Now just backsub
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For the last one Let $\displaystyle x=4\sec(\theta)$

I chose this because

$\displaystyle \left(4\sec(\theta)\right)^2-16=16\sec^2(\theta)-16=16\left(\sec^2(\theta)-1\right)=16\tan^2(\theta)$

5. Originally Posted by Isomorphism
$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}dx$

Use integration by parts:

$\displaystyle \int \frac{\ln(x)}{\sqrt{x}}\, dx = \ln x (2\sqrt{x}) - \int \frac{2\sqrt{x}}{x}\, dx = 2\sqrt{x} \ln x - 2\int \frac{1}{\sqrt{x}}\, dx = 2\sqrt{x} \ln x - 4\sqrt{x}+C$
Hmm...I didn't even think to use integration by parts

6. Originally Posted by Chris L T521
Hope that this makes sense!

--Chris
That's the problem, I don't understand how to even do those integrals for the first part. I have those equations, but I was lost.

The second part makes sense, so if I had the equation for a half circle, I'd just do $\displaystyle 2\pi \int f(x)dx$ f(x) being the equation for a circle?

7. Originally Posted by Debelius
That's the problem, I don't understand how to even do those integrals for the first part. I have those equations, but I was lost.

The second part makes sense, so if I had the equation for a half circle, I'd just do $\displaystyle 2\pi \int f(x)dx$ f(x) being the equation for a circle?
Not quite. For the second part, you would need to do $\displaystyle 2\pi\bar y {\iint\limits_R {dA}}$

Let me simplify the first part...

We can rewrite the equation $\displaystyle \bar x = \frac{{\iint\limits_R {x\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar x = \frac{1} {\text{Area of R}}{\iint\limits_R {x\,dA}}$ and $\displaystyle \bar y = \frac{{\iint\limits_R {y\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar y = \frac{1} {\text{Area of R}}{\iint\limits_R {y\,dA}}$

We can determine $\displaystyle \bar x$ two ways: by inspection or by evaluating the integral.

Since the semi-circle is symmetric about the y axis, we can determine that $\displaystyle \bar x=0$. Now if you didn't see that, then we would evaluate the equation for $\displaystyle \bar x$:

[Let me denote the radius of the circle as $\displaystyle r_0$ instead of $\displaystyle r$]...

$\displaystyle \bar x = \frac{1} {\text{Area of R}}{\iint\limits_R {x\,dA}}=\frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta$

Note how we are using polar coordinates to evaluate the integral.

We now see that:

\displaystyle \begin{aligned}\bar x &= \frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta \\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\cos\theta\right]_{r=0}^{r_0} \\ &= \frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}{r _0}^3\right)\int_0^{\pi}\cos\theta \,d\theta \\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\cdot 0 \\ &=\color{red}\boxed{0}\end{aligned}

Are you still with me?

Now let's find $\displaystyle \bar y$:

\displaystyle \begin{aligned} \bar y &=\frac{1}{\frac{1}{2}\pi{r_0}^2}{\iint\limits_ R {y\,dA}}\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\int_ 0^{r_0}(r\sin\theta)r\,dr\,d\theta\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\sin\theta\right]_{r=0}^{r_0}\,d\theta\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\int_0^{\pi}\sin\theta\,d\theta\end{ aligned}
\displaystyle \begin{aligned} \bar y&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3 }{r_0}^3\right)\bigg[-\cos\theta\bigg]_0^{\pi}\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{2}{3} {r_0}^3\right)\\ &=\color{red}\boxed{\frac{4r_0}{3\pi}}\end{aligned }

Thus, the centroid is $\displaystyle \color{red}\boxed{\bigg(0,\frac{4r_0}{3\pi}\bigg)}$

Now the second:

Using the theorem of Pappus, we can find this volume:

$\displaystyle V=2\pi \cdot \bar y \cdot \left(\text{Area of R}\right)$

$\displaystyle \therefore V=2\pi\cdot \left[\frac{4r_0}{3\pi}\right]\cdot \left[\frac{1}{2}\pi {r_0}^2\right]=\color{red}\boxed{\frac{4}{3}\pi{r_0}^3}$

We know this to be true because the volume of a sphere is $\displaystyle \frac{4}{3}\pi r^3$

I hope you can absorb all of this. If you have questions, feel free to ask!

--Chris

8. Originally Posted by Chris L T521
Not quite. For the second part, you would need to do $\displaystyle 2\pi\bar y {\iint\limits_R {dA}}$

Let me simplify the first part...

We can rewrite the equation $\displaystyle \bar x = \frac{{\iint\limits_R {x\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar x = \frac{1} {\text{Area of R}}{\iint\limits_R {x\,dA}}$ and $\displaystyle \bar y = \frac{{\iint\limits_R {y\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar y = \frac{1} {\text{Area of R}}{\iint\limits_R {y\,dA}}$

We can determine $\displaystyle \bar x$ two ways: by inspection or by evaluating the integral.

Since the semi-circle is symmetric about the y axis, we can determine that $\displaystyle \bar x=0$. Now if you didn't see that, then we would evaluate the equation for $\displaystyle \bar x$:

[Let me denote the radius of the circle as $\displaystyle r_0$ instead of $\displaystyle r$]...

$\displaystyle \bar x = \frac{1} {\text{Area of R}}{\iint\limits_R {x\,dA}}=\frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta$

Note how we are using polar coordinates to evaluate the integral.

We now see that:

\displaystyle \begin{aligned}\bar x &= \frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta \\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\cos\theta\right]_{r=0}^{r_0} \\ &= \frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}{r _0}^3\right)\int_0^{\pi}\cos\theta \,d\theta \\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\cdot 0 \\ &=\color{red}\boxed{0}\end{aligned}

Are you still with me?

Now let's find $\displaystyle \bar y$:

\displaystyle \begin{aligned} \bar y &=\frac{1}{\frac{1}{2}\pi{r_0}^2}{\iint\limits_ R {y\,dA}}\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\int_ 0^{r_0}(r\sin\theta)r\,dr\,d\theta\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\sin\theta\right]_{r=0}^{r_0}\,d\theta\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\int_0^{\pi}\sin\theta\,d\theta\end{ aligned}
\displaystyle \begin{aligned} \bar y&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3 }{r_0}^3\right)\left[-\cos\theta\right]_0^{\pi}\\ &=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{2}{3} {r_0}^3\right)\\ &=\color{red}\boxed{\frac{4r_0}{3\pi}}\end{aligned }

Thus, the centroid is $\displaystyle \color{red}\boxed{\bigg(0,\frac{4r_0}{3\pi}\bigg)}$

Now the second:

Using the theorem of Pappus, we can find this volume:

$\displaystyle V=2\pi \cdot \bar y \cdot \left(\text{Area of R}\right)$

$\displaystyle \therefore V=2\pi\cdot \left[\frac{4r_0}{3\pi}\right]\cdot \left[\frac{1}{2}\pi {r_0}^2\right]=\color{red}\boxed{\frac{4}{3}\pi{r_0}^3}$

We know this to be true because the volume of a sphere is $\displaystyle \frac{4}{3}\pi r^3$

I hope you can absorb all of this. If you have questions, feel free to ask!

--Chris
Hmm, great work as usual Chris, but uhm talk about quantam leap from integratin $\displaystyle \int\sin^2(\theta)\cos^3(\theta)d\theta$ to evaluating double integrals that need to be changed to polar coordinates to be able to be integrated by elementary terms.