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| Question | Answer |
| How do we find the volume of a cylinder? | π x r^2 x h |
| What is the formula involving the summation for volume of a cone? | π x lim(δ->0) x (b)Σ(a) x y^2 x δx |
| What do a and b represent in this formula? | The limits of the x-value |
| What does δx represent in this formula? | The thickness of the rectangles we are using to estimate the volume of the cone |
| Why should δx be as low as possible? | It will give a more accurate volume |
| Ways to improve our volume estimation are: | Increasing number of rectangles and increasing number of cylinders |
| Image to visualize it: | |
| This imagine represents: | Rectangles along x-axis rotation by 360 degrees (revolution) |
| As δx approaches 0, the sum becomes an.... | Integral |
| Another way to find the volume of a cone rotated by 360 degrees in the x-axis is by using the formula..... | π x b∫a (y^2) dx |
| We can also find the volume of a cone rotated 360 degrees in the y-axis by.... | π x b∫a (x^2) dx |
| 1. Region bounded between x^2 +5 between x=0 and x=4 and the x-axis is rotated 360 degrees about x-axis. Find the volume to 3sf | π x 4∫0 (x^2+5)^2 dx 4∫0 (x^2+5)^2 dx = 518.133 518.133 x π = 1627.76 3sf = 1630 |
| 2. Region bounded between x^2 +5 between y=6 and y=10 and the y-axis is rotated 360 degrees about the y-axis. Find the volume to 3sf | Rearrange y=x^2 +5 for x^2=y-5 so we can use formula π x b∫a (x^2) dx So 10∫6 (y-5)^2 dx = 12 π x 12 = 37.699 3sf = 37.7 |
| 3. Region bounded by y=3x^2 +1 between x=0 and x=2 and the y-axis is rotated by 360 degrees about the y-axis. Find the volume to 3sf | Find y boundaries: 3(0)^2 + 1 = 1 3(2)^2+1 = 13 y=3x^2 +1 rearranged = x^2 = (y-1)/3 Now: π x 13∫1 (y-1)/3 dx = 75.4 |
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