# Integration: Area under curves

• Apr 27th 2010, 12:59 AM
mag6
Integration: Area under curves
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.
• Apr 27th 2010, 01:50 AM
tom@ballooncalculus
whoops, sorry, used second axis for first.

However (dang! shouldn't have wiped it) H of I has a slip - should be $\displaystyle y = \pm\sqrt{x^2 - 7}$.
• Apr 27th 2010, 01:52 AM
HallsofIvy
Quote:

Originally Posted by mag6
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 $\displaystyle x^2- y^2= 7$ when $\displaystyle 16- y^2= 7$ or $\displaystyle y^2= 16- 7= 9$ so $\displaystyle y= \pm 3$.

If you draw the graph, you will see that the hyperbola crosses the x axis at $\displaystyle x= \pm\sqrt{7}$ and that the region between $\displaystyle x= \sqrt{7}$ 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, $\displaystyle y= \pm\sqrt{x^2+ 7}$.

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 $\displaystyle \pm\sqrt{x^2+ 7}$ and (x, 5) which are [tex]5+ \sqrt{x^2+ 7}[tex] and $\displaystyle 5- \sqrt{x^2+ 7}$. The area of a circle is $\displaystyle \pi r^2$ of course, so the area of those two circles is $\displaystyle \pi(5+ \sqrt{x^2+ 7})^2$ and $\displaystyle \pi(5-\sqrt{x^2+ 7})^2$ and the area of the "washer" between them is $\displaystyle \pi(5+ \sqrt{x^2+ 7})^2- \pi(5-\sqrt{x^+ 7})^2$. Multiplying that out, a lot ought to cancel.

The thickness of each washer is "dx" so the volume would be $\displaystyle \pi[(5+\sqrt{x^2+7})^2- (5- \sqrt{x^2+ 7})^2]dx$. Integrate that from $\displaystyle \sqrt{7}$ to 4.