Can you double check if there is a typo on the first function
x=y^2 => y=sqrt(x), ...(case 1) or is it
y=x^2 => x=sqrt(y).... (case 2)
In any case, the procedure is the same. I will work out the problem where y=x² (case 2).
Let f1(x)=3x (line), and
f2(x)=x² (curve)
Plot the two functions on a graph and determine the intersections.
Here is a plot (of case 2):
http://img854.imageshack.us/i/1299607814.png/
Call the intersections P1(0,0), P2(3,9).
For integration with respect to x, the limits of integration are x1,x2, and with respect to y, the limits are y1,y2.
1. integration with respect to x: (disk method).
Cut the region into vertical slices of thickness dx. When rotated about the x-axis, it will form a disk with a hole, of outside diameter f1(x), and inside diameter f2(x).
The volume of such a disk is therefore
dV=π(R²-r²)dx
=π(f1(x)²-f2(x)²)dx
The total volume is therefore
V=∫dV
=π∫(f1(x)²-f2(x)²)dx
2. Integrate with respect to y: (shell method)
Here, we need the inverse of each function, such that
for a given y=f1(x), x=g(y).
This can be obtained by solving for x in terms of y:
g1(y) = x = y/3, and
g2(y) = x = sqrt(y)
To find the volume, again we will be cutting slices, of thickness dy. However, when the slice is revolved around the x-axis, it becomes a shell (or hollow cylinder) of thickness dy, the height is (g2(y)-g1(y)), and the elemental volume
dV = surface area * thickness
= 2πrhdy
= 2π(y)(g2(y)-g1(y))dy
So the total volume
=∫dV
=2π∫y(sqrt(y)-y/3)dy
=2π∫(y^(3/2)-y²/3)dy
from y=0 to y=9
I get 162π/5 in both cases.
Consider the region bounded by the graphs of the equations x=y^2 and y=3x. Set up 2 integrals, one with respect to x and the other with respect to y, both of which compute the volume of the solid obtained by rotating this region about the x-axis and evaluate the integrals.
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