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What is the volume of the solid generated when S is revolved about the line y=3?
Full question:
Let S be the region enclosed by the graph of y=2x and y=2x^2 for [0,1]. What is the volume of the solid generated when S is revolved about the line y=3?
The answer is:
pi*integral( ( (3-2x^2)^2) - (3-2x)^2) ) dx ) on (0,1)
I just need to know how it got there. I know for sure that the order of subtraction in the integral should be top minus bottom, but I checked the graph and even plugged points to find that the top graph should be y=2x. Is it just a mistake on my part calculation wise, or did I copy the wrong answer?
This is not one of the do-my-homework-for-me type questions, I'm in genuine need of help for my AP test tomorrow. Please help, and thanks in advance.
1 Answer
- 7 years ago
To find the volume of a revolution you want to do pi * int([inner radius]^2 - [outer radius]^2)dx. Try drawing out the functions and thinking of what a revolution would look like. 3 minus the upper function will be your inner radius when you rotate and 3 minus the lower function will be the outer radius.
