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Mathematics 18 Online
OpenStudy (anonymous):

How do I find the volume generated by revolving a region enclosed by:

OpenStudy (anonymous):

\[y=2x ^{2}\] y=0 x=2

OpenStudy (anonymous):

Around the x-axis?

OpenStudy (anonymous):

Sorry, about the line y=8

OpenStudy (anonymous):

Since you're rotating about the line y=8, you want everything in terms of x. This means that the function is \[y=x \sqrt{2}\](We don't lose anything because the region is always positive.) Our limits of integration become x=0 (from y=0) and x=2 (from y=8). The equation for the area of a circle is A=pi*r^2. Our "height" (going horizontally now) will be a very small change in x, or dx. When we multiply an area by a height, we get a volume. This gives us the integral \[\int\limits_{0}^{2}\pi*(x \sqrt{2})^2dx\].Now it's just a relatively simply integral.

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