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a drinking glass has the shape of a truncated cone. If the internal radii of the base and the top are 3cm and 4cm respectively and the depth is 10cm, find a)by integration its capacity. b) the volume of water if the glass is filled with water to a depth of 5cm

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To find the volume of the cup filled only 5 cm, just change the bounds from 0 and 10 to 0 and 5
oh, I forgot to distribute that pi at the end, the answer would be 35pi/2

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Other answers:

Actually, I messed that up.... R(x) is supposed to be squared
If R(x) is a function to determine the distance from the axis of rotation and the outer edge of the cup, you can determine the volume of the cup using the Disc Method:\[\pi \int\limits_{a}^{b}[R(x)]^{2}dx\] So there are two things you need to determine: What is R(x) and what are the values for a and b (the integration bounds). To determine what R(x) is, imagine flipping the cup horizontally so the x axis runs through the center of the cup and the bottom of the cup is at x=0. The x axis is the axis of rotation, so find an equation to determine the y value of the edge of the cup for a given x|dw:1328438568479:dw| The y intercept is 1.5 and the slope is 0.5/10 or 5/100. That means R(x) is:\[R(x) = \frac{5}{100}x + 1.5\] The bounds, a and b, would be 0 and 10, because you're going from the bottom of the cup (which is at x=0) to the top of the cup (at x=10). Now you can integrate and solve:\[\pi \int\limits\limits\limits_{0}^{10}(\frac{5}{100}x+1.5)^{2}dx = \pi \int\limits\limits\limits_{0}^{10} \left(\frac{x^2}{16}+\frac{3x}{4}+\frac{9}{4}\right)dx = \pi \left[\frac{x^3}{48}+\frac{3x^{2}}{8}+\frac{9x}{4}\right]_{0}^{10}\] \[= \pi\left[\left(\frac{1000}{48} + \frac{300}{8}+\frac{90}{4}\right) - 0\right] = \frac{485}{6} \approx 80.8333 \]
And forgot to distribute the pi again... so it's about 253.945
thank you!!!!!!!
Actually that's wrong too, but I need to get some sleep or I'm just gonna keep messing up. Sorry :P
night night sleep tight so those neurotransmitters will be able to work tomorrow - then you can have another shot at this question thanks for all the help anyways, greatly appreciated!

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