Suppose a question asks for the volume of revolution about the x axis to be found on a piece of area enclosed between 2 graphs, where the area crosses the x-axis. In this case, the method involving the subtraction of two volumes does not seem to work as there is overlap due to the area crossing the x axis. How would this volume be calculated?
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3$\begingroup$ Use the revolution of max(f(x),-f(x)). $\endgroup$ – coffeemath Sep 2 '15 at 19:29
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$\begingroup$ @coffeemath What about for surface area please? $\endgroup$ – user636310 Mar 1 '19 at 12:37
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Suppose you have two functions $y_1$ and $y_2$ suitably well-behaved over an interval $[a,b]$ and you want the total volume created by rotating the area between the graphs about the $x$ axis. Suppose also that either graph could cross the $x$ or intersect with the other.
You would need to split the domain of integration into contiguous sub-intervals determined by the solutions of the equations $y_1=0$, $y_2=0$ and $|y_1|=|y_2|$.
If for any such interval $[p,q]$ we have $y_i\geq y_j\geq 0$ or $y_i\leq y_j\leq 0$, you need to calculate $$\pi\int_p^q(y_i^2-y_j^2)dx$$
And if for any such interval $[r,s]$ we have $y_i$ and $y_j$ have opposite parity, you need to determine which has larger magnitude. If this is $y_i$ for this interval, you only need $$\pi\int_r^s y_i^2 dx$$
Then you add the results.
answered Sep 2 '15 at 20:10
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1$\begingroup$ @Mitjackson the surface area of what exactly? Do you need to ask a new question, perhaps? $\endgroup$ – David Quinn Mar 2 '19 at 18:39
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$\begingroup$ Thanks David Quinn! Imagine the same question as this question by CowNorris but 'surface area' instead of 'volume'. Is the principle the same, namely that the overlap does not affect the value of the integral please? $\endgroup$ – user636310 Mar 3 '19 at 3:15
