Theorem of Pappus to find volume using the centroid

 
 
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Theorem of Pappus lets us find volume using the centroid and an integral

The Theorem of Pappus tells us that the volume of a three-dimensional solid object that’s created by rotating a two-dimensional shape around an axis is given by

???V=Ad???

where ???V??? is the volume of the three-dimensional object, ???A??? is the area of the two-dimensional figure being revolved, and ???d??? is the distance traveled by the centroid of the two-dimensional figure.

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Theorem of Pappus to find volume using the centroid of the plane region


 
 
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Finding volume of a right circular cone with Theorem of Pappus

Example

Use the Theorem of Pappus to find the volume of a right circular cone with radius ???r=6??? and height ???h=10???.


The Theorem of Pappus defines volume as ???V=Ad???. Before we can solve for volume we need to find the area of the triangle we’re revolving. Our shape, the right circular cone, can be described as a triangle rotated around an axis. The formula for area of a triangle is

???A=\frac12bh???

 
right circular cone
 

The base of the triangle will be the radius ???r=b=6???, and the height of the triangle will be ???h=10???.

???A=\frac12(6)(10)???

???A=30???

Next, we need to solve for distance, ???d???. Distance will involve the relationship of the triangle’s centroid and the rotation it experiences. In other words, ???d=2\pi{\overline{x}}??? where ???\overline{x}??? is the ???x???-coordinate of the centroid and ???2\pi??? refers to the fact that the object is being rotated around an axis. The equation for ???\overline{x}??? is

???\overline{x}=\frac{1}{A}\int^b_axf(x)\ dx???

Looking at this equation we realize we’re still missing ???f(x)???, which is the third side of the triangle, ???H???. There are different ways to do this, but we’ll use the pythagorean theorem and say that

???H^2=h^2+b^2???

where ???H??? is the hypotenuse, ???b??? is the base and ???h??? is the height.

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the volume of a three-dimensional solid object that’s created by rotating a two-dimensional shape around an axis is given by

V=Ad.

We’ll solve for ???H??? and then plug in the values we know.

???H=\sqrt{h^2+b^2}???

???H=\sqrt{(10)^2+(6)^2}???

???H=\sqrt{136}???

Remember that ???H=f(x)???, so

???f(x)=\sqrt{136}???

Now we can solve for ???\overline{x}???.

???\overline{x}=\frac{1}{30}\int^6_0x\sqrt{136}\ dx???

???\overline{x}=\frac{\sqrt{136}}{30}\int^6_0x\ dx???

???\overline{x}=\frac{\sqrt{136}}{30}\left(\frac{x^2}{2}\right)\bigg|^6_0???

???\overline{x}=\frac{\sqrt{136}}{30}\left[\frac{(6)^2}{2}-\frac{(0)^2}{2}\right]???

???\overline{x}=\frac{6\sqrt{34}}{5}???

Now we can solve for distance ???d=2\pi{\overline{x}}???.

???d=2\pi{\frac{6\sqrt{34}}{5}}???

???d=\frac{12\pi\sqrt{34}}{5}???

Finally, we can solve for volume using ???V=Ad???.

???V=(30)\left(\frac{12\pi\sqrt{34}}{5}\right)???

???V=72\pi\sqrt{34}???

Using the Theorem of Pappus, we know that the volume of a right circular cone with base radius ???r=6??? and height ???h=10??? is ???V=72\pi\sqrt{34}???.

 
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