Find the volume of the solid generated by revolving the region in the first quadrant that is above the parabola y= 4x^2 and below the parabola y= 45-x^2 about the y-axis
I just need help setting up the integral. V=∫[0,3]2pix(45-x^2)dx
What am I doing wrong?
Nevermind, I figured it out
V=∫[0,3]2pix(45-x^2-4x^2)dx
where do they hit?
4 x^2 = 45 - x^2
x^2 = 9
x = +/- 3
cylinder height = 45-x^2 - 4 x^2
= 45 - 5 x^2
cylinder thickness = dx
cylinder circumference = 2 pi x
so
V=∫[0,3]2 pi x(45-5x^2)dx
Yes, of course -4-1 = -5 :)
To find the volume of the solid generated by revolving a region around an axis, you need to use the method of cylindrical shells. However, it seems that you have mistakenly set up the limits of integration in your integral.
To correctly set up the integral, let's first find the points of intersection between the two parabolas. Set the two equations equal to each other and solve for x:
4x^2 = 45 - x^2
5x^2 = 45
x^2 = 9
x = ±3
The region in the first quadrant that is bounded by the parabolas is from x = 0 to x = 3. Therefore, the correct limits of integration for the integral should be [0, 3].
Now, let's consider a thin vertical strip at x with thickness dx. When this strip is rotated around the y-axis, it forms a cylindrical shell. The height of this shell is given by the difference in the y-values of the two parabolas at x, which is (45 - x^2) - 4x^2 = 45 - 5x^2. The circumference of the shell is given by 2πx.
Therefore, the volume of a single cylindrical shell is given by dV = 2πx * (45 - 5x^2) * dx.
To find the total volume, integrate the expression for dV over the range [0, 3]:
V = ∫[0,3] 2πx * (45 - 5x^2) dx
Now, you can evaluate this integral to find the volume of the solid generated by revolving the given region around the y-axis.