Find the volume of the solid generated by revolving the region bounded by
y=e^(5/2x),y=0,x=1, and x=5
about the x-axis.
ok ok ok
I did one. Can you follow my logic and try the others? They're all done the same way. Decide the limits of integration, and then use
discs: v = ∫πr^2 dx
shells: v = ∫2πrh dx
or dy, depending on how you want to integrate. Just decide on r and h, and plug them in.
To find the volume of the solid generated by revolving the region bounded by the given curves about the x-axis, we can use the method of cylindrical shells.
To do this, we need to set up an integral that represents the volume of each cylindrical shell. We can consider each infinitesimally thin shell as a hollow cylinder with thickness "dx", height "y", and radius "x".
The radius of each cylindrical shell is equal to the x-coordinate of the curve at a given x-value. In this case, the radius is "x". The height of each shell is given by the difference between the two curves at each value of x. Therefore, the height is "e^(5/2x)" (upper curve) minus 0 (lower curve), which simplifies to "e^(5/2x)".
To find the limits of integration, we look at the given bounds for x: x=1 and x=5. So, we integrate with respect to x over the interval [1,5].
The volume of each cylindrical shell is given by the formula:
dV = 2πxy dx
Therefore, the integral to find the total volume is:
V = ∫ from 1 to 5 (2πxe^(5/2x)) dx
Now, we can evaluate this integral to find the volume of the solid.