Keesha Q.

asked • 06/18/20

Hi i need some help with this question

A region in the first quadrant is bounded above by the curve y = tanh x, below by the x-axis on the left by the  y-axis and on the right by the line x = ln 7.  Find the volume of the solid generated by revolving the region about the x-axis.

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Doug C. answered • 06/18/20

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David M.

I would like to back up somewhat to understand where the formula (Pi)*{integral of [(tan(x))^2]dx from x=0 to x=LN 7} comes from. because understanding that should give insights of how to do similar types of problems that you do you may not have seen before. If the student knows how the formula came about, following can be ignored. When you have a rotation on a function creating a solid, you want to divide the solid up into discs having a radius that is same as the value of the function minus axis of rotation, which axis is y=0 in our case, so the radius in our case is the value of the function itself. The volume of each disk is the area of circular face, which is (Pi)r^2 in general and (Pi)(f(x))^2= (Pi)(tan(x))^2 in our case. times the disc's thickness, which we will take to be dx as in infinitesimally small change in x. We then add up the volume of all of our discs to get the volume of the whole solid, which, because each disc's thickness is infinitesimally small, dx, translates to the integral of (Pi)(r^2)dx from x=0 to x=LN 7 =integral of (Pi)(r^2)dx from x=0 to x=LN 7 =integral of (Pi)[(tan(x))^2]dx from x=0 to x=LN 7 and because the integral of a constant times a function with respect to x equals the constant times the integral of the function with respect to x, then we have (Pi)*{integral of [(tan(x))^2]dx from x=0 to x=LN 7}
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06/20/20

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