Consider the following region between the graph of the equations y=4x and y=x^3. Set up an integral to compute the volume of the solid obtained by rotating this region about the line x=-1.
2007-03-17 04:10:39 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
it would be the integral of pi (4x-1)^2 - (x^3-1)^2 + the integral of pi (X^3-1)^2 - (4x-1)^2
You foil and it would be the inegral of pi 16x^2-8x-x^6+2x^3.
You foil the other and get X^6-2x^3-16x^2+8x^2
Integrate and get: pi (16x^3/3)-(8x^3/3)-(x^7/7)+(X^4/2)
You integrate the other and get pi (x^7/7)-(x^4/2)-(16x^3/3)+(8x^3/3)
Substitute the upper bound and the lower bound for both(-2, 0) and (0,2), subtract those two. Then add them both together and you get the volume!!!!
2007-03-17 04:32:50 · answer #1 · answered by elphaba_of_georgia 3 · 0⤊ 0⤋
Please note that since this is a volume generated by revolution parallel to the y axis it can't be done by an integration with respect to x as the previous person suggests. This would give a volume revolved parallel to the x axis.
This line and curve intersect at three points (0,0), (2,8) and
(-2,-8). Therefore the finite region between them consists of two parts one in the first quadrant and one in the third quadrant. If you only want the volume formed by the first quadrant part rotated about the line x = -1 then this is straightforward. It is done as the difference of two integrals of the form pi*int(x + 1)^2dy with limits of 8 and 0. You must first convert each function to give x in terms of y. These are
x = y^(1/3) and x = y/4. Thus you need
pi*int((y^(1/3) + 1)^2)dy - pi*int(((y/4) + 1)^2dy)
between y = 8 and y = 0.
If you want to involve the other region in the third quadrant things get very difficult as the line x = -1 cuts this region.
2007-03-17 11:41:23 · answer #2 · answered by mathsmanretired 7 · 0⤊ 0⤋
What are the bounds of your integral? or do you want the volume from -1 to infinity?
2007-03-17 11:25:14 · answer #3 · answered by Dave 6 · 0⤊ 1⤋