Exercise

Find the length of the curve.

    1. MATHansMATH

    2. MATHansMATH

    3. MATHans$:6] $

    4. MATHans$:\frac{17}{6}]$

    5. MATHans$:\frac{22}{3}]$

    6. MATHansMATH

    7. MATHans$:\ln (1+\sqrt{2})]$

    8. MATHansMATH

    9. MATHansMATH

    10. MATHans$:\frac{3}{2}]$

  2. Find the surface area generated when the curve is revolved about the x-axis.

    1. $y=x^{2},[1,3]$ \lbrack ans:MATH]

    2. $y=x+1,[0,2]\qquad $[ans:8$\pi \sqrt{2}$]

    3. MATH[ans:$\frac{49\pi }{3}$]

    4. MATH[ans:$\frac{47}{16}\pi $]

CENTROID OF A REGION; PAPPUS'S THEOREM ON VOLUMES

The Centroid of a Region

Suppose that we have a thin distribution of matter, a plate, laid out in the $xy$-plane in the shape of some region $\Omega $. If the mass density of the plate varies from point to point, then the determination of the center of mass of the plate requires double integration. If, however, the mass density of the plate is constant throughout $\Omega $, then the center of mass of the plate depends only on the shape of $\Omega $ and falls on a point that we call the centroid of $\Omega $. Unless $\Omega $ has a very complicated shape, we can calculate the centroid of $\Omega $ by ordinary one-variable integration.

We will use two guiding principles to find the centroid of a region $\Omega $. The first is obvious. The second we take from physics.

Principle 1: Symmetry. If the region has an axis of symmetry, then the centroid MATH lies somewhere along that axis. (It follows from Principle 1 that, if the region has a center, then that center is the centroid.)

Principle 2: Additivity. If the region, having area $A,$ consists of a finite number of pieces with areas MATH and centroids MATH , then
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and
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We are now ready to bring the techniques of calculus into play. Let's denote the area of $\Omega $ by $A$. The centroid MATH of $\Omega $ can be obtained from the following formulas:
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To derive these formulas we choose a partition
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of $[a,b].$ This breaks up $[a,b]$ into $n$ subintervals. Choosing $x_{i}^{\ast }$ as the midpoint of $[x_{i-1},x_{i}]$, we form the midpoint rectangles $R_{i}$.The area of $R_{i}$ is MATH and the centroid of $R_{i}$ is its center MATH By Principle 1, the centroid MATH of the union of all these rectangles satisfies the following equations:
MATH

MATH

(Here $A_{P}$ represents the area of the union of the $n$ rectangles.) As MATH, the union of rectangles tends to the shape of $\Omega $ and the equations we just derived tend to the formulas
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Problem 1

Find the centroid of the quarter-disc
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Solution.

The quarter-disc is symmetric about the line $x=y.$ We know therefore that MATH Here
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Since $A=\U{bc}\pi r^{2},$
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The centroid of the quarter-disc is the point
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Problem 2

Find the centroid of the right triangle with vertices $(0,0),(b,0),(0,h).$

Solution

There is no symmetry that we can use here. The hypotenuse lies on the line
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Hence
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and
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Since $A=\U{bd}bh,$ we have
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and
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Let the region $\Omega $ be between the graphs of two continuous functions $f $ and $g.$ In this case, if $\Omega $ has area $A$ and centroid MATH then
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Proof.

Let $A_{f}$ be the area below the graph of $f$ and let $A_{g}$ be the area below the graph of $g.$ Then in obvious notation


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and
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Therefore
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and
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Problem 3

Find the centroid of the region bounded by
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Solution. Here there is no symmetry we can appeal to. We must carry out the calculations.
MATH

MATH

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Therefore
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Pappus's Theorem on Volumes

All the formulas that we have derived for volumes of solids of revolution are simple corollaries to an observation made by a brilliant, ancient Greek, Pappus of Alexandria (circa 300 A.D.).

PAPPUS'S THEOREM ON VOLUMES

A plane region is revolved about an axis that lies in its plane. If the region does not cross the axis, then the volume of the resulting solid of revolution is the area of the region multiplied by the circumference of the circle described by the centroid of the region:
MATH
where $A$ is the area of the region and $\overline{R}$ is the distance from the axis to the centroid of the region.

As special cases of Pappus's Theorem we have

1. The Washer Method Formula. If MATH and if the region bounded by $y=f(x)$ and $y=g(x)$ is revolved about the $x$-axis, the resulting solid has volume
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2. The Shell Method Formula. If MATH and if the region bounded by $y=f(x)$ and $y=g(x)$ is revolved about the $y$-axis, the resulting solid has volume
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Note that
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and
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Applications of Pappus's Theorem

Problem 4

Find the volume of the solid generated by revolving this region bounded by
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about the line $y=5.$

Solution

Here MATH and $A=\frac{4}{3}$. Hence
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Problem 5

Find the volume of the doughnut (called torus in mathematics) generated by revolving the circular disc
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about (a) the $x$-axis, (b) the $y$-axis.

Solution

The centroid of the disc is the center $(a,b).$ This lies $b$ units from the $x$-axis and $a$ units from the $y$-axis. The area of the disc is $\pi c^{2}. $ Therefore

(a)
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(b)
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Problem 6

Find the centroid of the half-disc
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by appealing to Pappus's theorem.

Solution. Since the half-disc is symmetric about the $y$-axis, we know that $\overline{x}=0.$ All we need to find is $\overline{y}$.

If we revolve the half-disc about the x-axis, we obtain a solid ball of volume MATH. The area of the half-disc is $\U{bd}\pi r^{2}.$ By Pappus's theorem
MATH

Simple division gives $\overline{y}=$ $\frac{4r}{3\pi }$.

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