Find the volume of the solid obtained by rotating the region bounded by y=30x-5x^2 and y=0 about the y-axis! Please Help! I can't get the answer no matter how hard I try!
2007-04-27 09:25:43 · 4 answers · asked by garrett m 1 in Science & Mathematics ➔ Mathematics
A little graphing goes a long way here. This is an upside down parabola with a peak at y = 45 and intersecting the x axis at x = 0 and x = 6.
By symmetry, the centroid of the area lies on the line x = 3. Rotating about the y axis will cause the centroid to move a distance c = 2πr = 2*π*3 = 6π units.
The required volume is then the area of the cross section times 6π. Ac = ∫(30x - 5x²)dx from 0 to 6 = 15x² - (5/3)x³│= 180 units²
Finally, V = 6πAc = 3392.92 units³
2007-04-27 09:51:56 · answer #1 · answered by Steve 7 · 0⤊ 0⤋
First find the zeros of the function.
y = 30x - 5x² = 5x(6 - x) = 0
x = 0,6
Integrate using the cylindrical shells method. Integrate over the interval x = 0 to x = 6.
â«2Ïxy dx = 2Ïâ«x(30x - 5x²)dx
= 10Ïâ«(6x² - x³)dx
= 10Ï[2x³ - x^4/4) | [Evaluated from x = 0 to x = 6]
= 10Ï[2*216 - 1296/4] = 10Ï[432 - 324] = 1080Ï
2007-04-27 16:43:44 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
well its been a while since Ive done this but if i remember correctly will be to find the area under the curve of y=30x-5x^2
and then transform into cylindrical cordinates , then take the integral from 0 to 2 * pi
Sorry, I cant remeber everything but this is a start
2007-04-27 16:39:15 · answer #3 · answered by dragongml 3 · 0⤊ 0⤋
540
**edit**
Oops, forgot the 2*pi.
2007-04-27 16:40:28 · answer #4 · answered by sweetwater 7 · 0⤊ 0⤋