3.3 Applications of Double Integrals

(a) $6\pi$ (b) $\half ab (a+b)$

(a) $\dblInt_D x\ \dee{x}\,\dee{y}=0$ because $x$ is odd under $x\rightarrow-x$, i.e. under reflection about the $y$–axis, while the domain of integration is symmetric about the $y$–axis. $\dblInt_D 3\ \dee{x}\,\dee{y}$ is the three times the area of a half disc of radius $2$. So, $\dblInt_D(x+3)\dee{x}\,\dee{y}=3\times \half\times\pi 2^2=6\pi$.

(s) $\dblInt_R x\ \dee{x}\,\dee{y}/\dblInt_R \dee{x}\,\dee{y}$ is the average value of $x$ in the rectangle $R$, namely $\frac{a}{2}$. Similarly, $\dblInt_R y\ \dee{x}\,\dee{y}/\dblInt_R \dee{x}\,\dee{y}$ is the average value of $y$ in the rectangle $R$, namely $\frac{b}{2}$. $\dblInt_R \dee{x}\,\dee{y}$ is area of the rectangle $R$, namely $ab$. So,

and $\dblInt_R (x+y)\dee{x}\,\dee{y}=\half ab (a+b)$.

$\bar x=0$ and $\bar y = \frac{5}{7}$.

Here is a sketch of $D$.

By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $D$. That is,

By symmetry under reflection in the $y$–axis, we have $\bar x=0$. So we just have to determine $\bar y$. We'll evaluate the integrals using vertical strips as in the figure above. Looking at that figure, we see that

So the denominator is

and the numerator of $\bar y$ is

All together, $\bar x=0$ and

(a)

(b) $\text{mass} = \frac{4\pi}{3} - 2\ln\big(2+\sqrt{3}\big)$

(c) $\bar x = \frac{2\sqrt{3}- \ln(2+\sqrt{3})} {\frac{4\pi}{3} - 2\ln(2+\sqrt{3})} \approx 1.38$, $\bar y=0$.

(a) Here is a sketch of $R$.

(b) Considering that

we'll use polar coordinates.

Observe that the line $x=1$ and the circle $x^2+y^2=4$ intersect when

and that the polar coordinates of the point $(x,y)=\big(1,\sqrt{3}\big)$ are $r=\sqrt{x^2+y^2}=2$ and $\theta=\arctan\frac{y}{x}=\arctan \sqrt{3} =\frac{\pi}{3}$. Looking at the sketch

we see that, on $R$,

So the mass is

(c) By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $R$. That is,

By symmetry under reflection in the $x$–axis, we have $\bar y=0$. So we just have to determine $\bar x$. The numerator is

All together, $\bar y=0$ and

$\bar x = \frac{10}{3\pi+8} \approx 0.57$

Let's call the plate $\cP$. By definition, the $x$–coordinate of its centre of mass is

Here is a sketch of the plate.

The cardiod is given to us in polar coordinates, so let's evaluate the integrals in polar coordinates. Looking at the sketch above, we see that, on $\cP$,

So the two integrals of interest are

and

All together

For an efficient, sneaky, way to evaluate
$\int_0^{\pi/2}\sin^2\theta\ \dee{\theta}$, see Remark 3.3.5 in the CLP-3 text.

$\bar x = \bar y =\frac{4}{3\pi}$

Call the plate $P$. By definition, the centre of mass is $(\bar x, \bar y)$, with $\bar x$ and $\bar y$ being the weighted averages of the $x$ and $y$–coordinates, respectively, over $P$. That is,

with $\rho(x,y)=k$. Here is a sketch of $P$.

By symmetry under reflection in the line $y=x$, we have $\bar y=\bar x$. So we just have to determine

The denominator is just one quarter of the area of circular disk of radius $1$. That is, $\dblInt_P \dee{A}=\frac{\pi}{4}$. We'll evaluate the numerator using polar coordinates as in the figure above. Looking at that figure, we see that

As $\dee{A}=r\,\dee{r}\,\dee{\theta}$, and $x=r\cos\theta$, the numerator

All together

$\frac{6}{5}$

Here is a sketch of $R$.

Note that

By definition, the $y$–coordinate of the center of mass of $R$ is the weighted average of $y$ over $R$, which is

On $R$,

Thus

For both $n=2$ and $n=3$, we have

So

$\frac{2}{3}$

By the definition given in the statement with $(a,b)=(0,0)$, the average is

The denominator $A(D) = \pi$. We'll use polar coordinates to evaluate the numerator.

So the average is

(a) $\frac{3\pi}{4}$ (b) $-\frac{1}{6}$

Note that $x^2+y^2=x$ is equivalent to $\left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4}$, which is the circle of radius $\frac{1}{2}$ centred on $\left(\frac{1}{2},0\right)$. Let's call the crescent $\cC$ and write

so that

meaning that $\cC$ is the disk $D$ with the “hole” $H$ removed. Here is a sketch.

(a) As $D$ is a disk of radius $1$, it has area $\pi$. As $H$ is a disk of radius $\nicefrac{1}{2}$, it has area $\nicefrac{\pi}{4}$. As $\cC$ has density $1$,

(b) Recall that, by definition, the $x$–coordinate of the centre of mass of $\cC$ is the average value of $x$ over $\cC$, which is

We have already found that $\dblInt_\cC \dee{A}=\frac{3\pi}{4}$. So we have to determine the numerator

As $x$ is an odd function and $D$ is invariant under $x\rightarrow -x$, $\dblInt_D x\,\dee{A}=0$. So we just have to determine $\dblInt_H x\,\dee{A}$. To do so we'll work in polar coordinates, so that $\dee{A} = r\,\dee{r}\,\dee{\theta}$. In polar coordinates $x^2 + y^2 = x$ is $r^2 =r\cos\theta$ or $r=\cos\theta$. So, looking at the figure above (just before the solution to part (a)), on the domain of integration,

So the integral is

So all together

Try using polar coordinates.

$4\sqrt{2} -\sqrt{2}\pi\approx 1.214$

The domain is pictured below.

The two circles intersect when $x^2+y^2=2$ and

In polar coordinates $x^2+y^2=2$ is $r=\sqrt{2}$ and $x^2+(y-1)^2=x^2+y^2-2y+1=1$ is $r^2-2r\sin\theta=0$ or $r=2\sin\theta$. The two curves intersect when $r=\sqrt{2}$ and$\sqrt{2}=2\sin\theta$ so that $\theta=\frac{\pi}{4}$ or $\frac{3}{4}\pi$. So

and, as the density is $\frac{2}{r}$,

(a) $\frac{1}{3}(b-a\,,\,c)$ (b) See the solution.

(a) The side of the triangle from $(-a,0)$ to $(0,c)$ is straight line that passes through those two points. As $y=0$ when $x=-a$, the line must have an equation of the form $y=K(x+a)$ for some constant $K$. Since $y=c$ when $x=0$, the constant $K=\frac{c}{a}$. So that the equation is $y=\frac{c}{a}(x+a)$. has equation $cx-ay=-ac$. Similarly the side of the triangle from $(b,0)$ to $(0,c)$ has equation $y=\frac{c}{b}(b-x)$. The triangle has area $A=\frac{1}{2}(a+b)c$. It has centre of mass $(\bar x,\bar y)$ with

To evaluate the integrals we'll decompose the triangle into vertical strips as in the figure

(b) The midpoint of the side opposite $(-a,0)$ is $\frac{1}{2}\big[(b,0)+(0,c)\big]=\frac{1}{2}(b,c)$. The vector from $(-a,0)$ to $\frac{1}{2}(b,c)$ is $\frac{1}{2}\llt b,c\rgt-\llt-a,0\rgt =\llt a+\frac{b}{2},\frac{c}{2}\rgt$. So the line joining these two points has vector parametric equation

The point $(\bar x,\bar y)$ lies on this line since

Similarly, the midpoint of the side opposite $(b,0)$ is $\frac{1}{2}(-a,c)$. The line joining these two points has vector parametric equation

The point $(\bar x,\bar y)$ lies on this line too, since

It is not really necessary to check that $(\bar x,\bar y)$ lies on the third median, but let's do it anyway. The midpoint of the side opposite $(0,c)$ is $\frac{1}{2}(b-a,0)$. The line joining these two points has vector parametric equation

The point $(\bar x,\bar y)$ lies on this median too, since