6GVolumeOfSolids

Volume of Solids of Revolution

Consider the area bounded by math . \quad \cdot \;\; \text{ and the rule: } y=\dfrac{x}{2} math
 * the x-axis,
 * the line x = 4

See diagram on the right. The shaded region is clearly a triangle.

We are now going to rotate the shaded region in 3 dimensions around the x-axis.

Rotating our triangle with the x-axis as the pivot would sweep out a 3D solid shape -- in this case a cone lying on its side.

We call this shape a __**Solid of Revolution**__.

(Because we made it by revolving the shaded region and it is solid.)

To find the **__Volume__** of this solid shape, we slice it into thin disks.

Each disk is __approximately__ a cylinder, with a radius (r) equal to the y-value of the function at that point.

The height of the cylinder is the width of the slice (h).

math \text{Volume(disk) } \approx \pi r^2 \,h math

but r = y-value and if we make h really small (h → 0) we can replace it with **dx**

As h → 0, the volume of our slice approaches the exact volume of a cylinder.

math \text{Volume(disk) } = \pi y^2 \, dx math

To get the volume of the entire solid, we sum all the disks together from **x = a** to **b**

math \text{Volume(solid) } = \displaystyle{\int\limits_{x=a}^{x=b} \pi y^2 \; dx} math

Example

Find the Volume of the Solid of Revolution formed by rotating y = f(x) around the x-axis between x = 0cm and 4cm. math \text{Where } \, y = \dfrac{x}{2} math

Use the rule: math \text{Volume(solid) } = \displaystyle{ \int\limits_{x=a}^{x=b} \pi y^2 \; dx} math . math \text{Volume } = \displaystyle{\int\limits_{x=0}^{x=4} \pi \left( \dfrac{x}{2} \right)^2 \; }dx math . math . \qquad \quad = \displaystyle{\int\limits_{x=0}^{x=4} \dfrac{\pi x^2}{4} \; }dx math . math . \qquad \quad = \left[ \dfrac{\pi}{4} \centerdot \dfrac{x^3}{3} \right]_{x=0}^{x=4} math . math \\ . \qquad \quad = \dfrac{64 \pi}{12} - 0 \\. \\ . \qquad \quad = \dfrac{16 \pi}{3} \; \text{ cubic units} \\. \\ . \qquad \quad \approx 16.75 \, cm^3 math

**See also:** www.mathsonline.com.au Y12Advanced --> Calculus --> Integration --> Lesson 9

Rotating around the y-axis

If the solid of revolution is formed by rotating a region around the y-axis, the rule is the same only with x and y swapped.

math . \qquad \text{Volume(solid) } = \displaystyle{ \int\limits_{y=a}^{y=b} \pi x^2 \; dy} math

Example Find the volume of revolution formed by rotating **//y = x 2 //** around the y-axis between y = 2cm and y = 4cm. math \text{Since } y=x^2 \; \Rightarrow \; x=\sqrt{y} math

Use the rule math . \qquad \text{Volume(solid) } = \displaystyle{ \int\limits_{y=a}^{y=b} \pi x^2 \; dy} math

math . \qquad \text{Volume } = \displaystyle{\int\limits_{y=2}^{y=4} \pi y \; }dy math . math . \qquad \qquad = \left[ \dfrac{\pi y^2}{2} \right]_{y=2}^{y=4} math . math . \qquad \quad = \dfrac{16\pi}{2} - \dfrac{4\pi}{2} math . math \\ . \qquad \quad = 8\pi - 2\pi \\. \\ . \qquad \quad = 6\pi \; \text{ cubic units} \\. \\ . \qquad \quad \approx 18.85 \, cm^3 math

Volume of a solid formed by rotating around the axis a region which is the area between two curves.

The rule is math \text{Volume(solid) } = \displaystyle{ \int\limits_{x=a}^{x=b} \pi \left( f^2 - g^2 \right) \; }dx math Where **f** is the upper curve and **g** is the lower curve.

Example Find the volume of the solid of revolution formed by rotating around the x-axis, the area between **//f(x) = –x 2 + 4x//** and **//g(x) = x 2 – 4x + 6//**.

Sketching the two curves reveals that **f(x)** is the __upper__ curve and **g(x)** is the __lower__ curve.

No limits are stated, so we are expected to use the intersection of the curves. Using algebra, or the calculator, we find the curves intersect at **x = 1** and **x = 3**.

If we rotate the shaded region around the x-axis, it would produce a ring donut shape. {except it would have a ridge running around the sides of the donut}

The volume of this shape could be calculated by: math V = \displaystyle{ \int\limits_{x=a}^{x=b} \pi f^2 \; }dx - \displaystyle{ \int\limits_{x=a}^{x=b} \pi g^2 \; }dx math

This simplifies to math V = \displaystyle{ \int\limits_{x=a}^{x=b} \pi \left( f^2 - g^2 \right) \; }dx math

Note that the functions must be squared BEFORE subtracting!!

math \text{Volume } = \displaystyle{ \int\limits_{x=a}^{x=b} \pi \left( f^2 - g^2 \right) \; }dx math . math . \qquad \quad = \displaystyle{ \int\limits_{x=1}^{x=3} \pi \left( (-x^2+4x)^2 - (x^2-4x+6)^2 \right) \; }dx math . math . \qquad \quad = \pi \displaystyle{ \int\limits_{x=1}^{x=3} \left( x^4-8x^3+16x^2 \right) - \left(x^4-8x^3+28x^2-48x+36 \right) \; }dx math . math . \qquad \quad = \pi \displaystyle{ \int\limits_{x=1}^{x=3} -12x^2+48x-36 \; }dx math . math . \qquad \quad = \pi \left[ \dfrac{-12x^3}{3} + \dfrac{48x^2}{2} -36x \right]_{x=1}^{x=3} math . math . \qquad \quad = \pi \left[ -4x^3 + 24x^2 -36x \right]_{x=1}^{x=3} math . math . \qquad \quad = \pi \left[ \left( -4 \times 3^3 + 24 \times 3^2 -36 \times 3 \right) - \left( -4 \times 1^3 + 24 \times 1^2 -36 \times 1 \right) \right] math . math \\ . \qquad \quad = \pi \left[ 0 + 16 \right] \\. \\ . \qquad \quad = 16\pi \; \text{ cubic units} math

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