whoops, sorry, used second axis for first.
However (dang! shouldn't have wiped it) H of I has a slip - should be .
Set up but do not evaluate the an integral for the volume of the solid obtained by rotating the region bounded by the given curves about specified axis
1. x^2 -y^2 =7 , x=4 about y=5
2. y= (x-2)^4 , 8x-y=16 about x=10
in the second one i know that they are intersecting at (2,0) and (4,16) but I need help setting up the integral for both of them.
For the first, x= 4 intersects when or so .
If you draw the graph, you will see that the hyperbola crosses the x axis at and that the region between and x= 4 is the only bounded area. The line y= 5 is well above that region so rotating around it will give a "doughnut" shaped figure. The line x= constant will cross the region in two places, .
If you draw a vertical line representing x= constant, you will see that the region rotates around the axis in a "washer" which is simplest to represent sa the area between two circles. The radius of each circle is the distance from the point and (x, 5) which are [tex]5+ \sqrt{x^2+ 7}[tex] and . The area of a circle is of course, so the area of those two circles is and and the area of the "washer" between them is . Multiplying that out, a lot ought to cancel.
The thickness of each washer is "dx" so the volume would be . Integrate that from to 4.