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| − | In [[calculus]], solids are formed by rotating a curve around an axis and integrating to find the volume. Typically the [[integration]] is of slices cut vertically to the axis of the rotation that formed the solid. Those slices are then integrated from one end of the solid to the other. | + | In [[calculus]], '''solids''' are formed by rotating a curve around an axis and integrating to find the volume. Typically the [[integration]] is of slices cut vertically to the axis of the rotation that formed the solid. Those slices are then integrated from one end of the solid to the other. |
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| | == Example == | | == Example == |
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| − | Consider the region in the first quadrant that has an upper bound of <math>y = \sqrt 2</math> and a lower bound of <big><math>y = (\sec{x})(\tan{x})</math></big>, and bounded on the left side by the ''y-axis''. Find the volume of the solid formed by rotating the region about the line <math>y = \sqrt 2</math>. | + | Consider the region in the first quadrant that has an upper bound of <math>y = \sqrt 2</math> and a lower bound of <big><math>y = (\sec{x})\,(\tan{x})</math></big>, and bounded on the left side by the ''y-axis''. Find the volume of the solid formed by rotating the region about the line <math>y = \sqrt 2</math>. |
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| | [[Calculus]] provides an elegant way to determine the volume of this solid. First, find where the curves intersect in order to ascertain the end-point of the integration. The boundaries intersect where | | [[Calculus]] provides an elegant way to determine the volume of this solid. First, find where the curves intersect in order to ascertain the end-point of the integration. The boundaries intersect where |
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| | on the other side. | | on the other side. |
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| − | We are not ready to find the volume. Note first that: | + | We are now ready to find the volume. Note first that: |
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| | :<math> dV = {\pi}r^2dx</math> | | :<math> dV = {\pi}r^2dx</math> |
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| | :<math>V = \int_0^\frac{\pi}{4} {\pi}r^2\,dx</math> | | :<math>V = \int_0^\frac{\pi}{4} {\pi}r^2\,dx</math> |
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| − | [[category:mathematics]] | + | The next insight is to express ''r'' in terms of ''x''. The variable ''r'' is the distance of the boundary from the axis about which it is rotated: |
| − | [[category:calculus]] | + | |
| | + | :<math>r = \sqrt 2 - \sec{x}\,\tan{x}</math> |
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| | + | The volume then becomes: |
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| | + | :<math>V = \int_0^\frac{\pi}{4} {\pi}(\sqrt 2 - \sec{x}\,\tan{x})^2\,dx</math> |
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| | + | :<math>V = \int_0^\frac{\pi}{4} {\pi}(2 - 2\sqrt2\sec{x}\,\tan{x} + (\sec{x}\,\tan{x})^2\,dx</math> |
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| | + | The only challenging part of this [[integral]] is the last term, which must be [[integrated by parts]]: |
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| | + | :<math>\int_0^\frac{\pi}{4}(\sec{x}\,\tan{x})^2\,dx</math> |
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| | + | :<math> = \int_0^\frac{\pi}{4}\sin{x}\,(\frac{\sin{x}}{\cos^4{x}})\,dx</math> |
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| | + | :<math> = \frac{\sin{x}}{3\cos^3{x}} - \int_0^\frac{\pi}{4}\frac{\sec^2{x}}{3},\,dx</math> |
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| | + | Recall that: |
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| | + | :<big><math>\int\sec^2{x}\,dx = \tan{x}</math></big> |
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| | + | and the solution to the overall integral is easy to obtain. |
| | + | [[Category:Mathematics]] |
| | + | [[Category:Calculus]] |