Answer:
40π
Explanation:
First, find the limits (intersections).
5x² = 30x − 10x²
15x² − 30x = 0
x² − 2x = 0
x (x − 2) = 0
x = 0 or 2
Within this interval, 30x − 10x² is greater than 5x².
Dividing the volume into cylindrical shells, the volume of each shell is:
dV = 2π r h t
dV = 2π x (30x − 10x² − 5x²) dx
dV = 2π x (30x − 15x²) dx
dV = 30π (2x² − x³) dx
The total volume is the sum (integral):
V = ∫ dV
V = ∫₀² 30π (2x² − x³) dx
V = 30π ∫₀² (2x² − x³) dx
V = 30π (⅔ x³ − ¼ x⁴)|₀²
V = 30π (⅔ 8 − ¼ 16)
V = 30π (16/3 − 4)
V = 10π (16 − 12)
V = 40π
Following are the calculation to the volume:
Given:
[tex]y = 5x^2\\\\ y = 30x - 10x^2[/tex]
To find:
Volume=?
Solution:
Equating the curves:
[tex]5x^2 = 30x - 10x^2\\\\15x^2 - 30x = 0\\\\x^2 - 2x = 0\\\\x (x - 2) = 0\\\\x = 0\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x- 2=0\\\\x = 0\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x= 2\\\\[/tex]
Volume generated by y-axis [tex]= \int_{a}^{b} 2\pi r h \ dx\\\\[/tex]
[tex]= \int^{2}_{0}\ 2 \pi x (30x - 10x^2 - 5x^2) \ dx\\\\= 2 \pi \int^{2}_{0} x (30x - 15x^2) \ dx\\\\= 30 \pi \int^{2}_{0} (2x^2 - x^3) dx\\\\= 30 \pi ( \frac{2x^3}{3} - \frac{x^4}{4})^{2}_{0}) \\\\= 30 \pi ( \frac{2(2)^3}{3} - \frac{(2)^4}{4}-0)] \\\\= 30 \pi [( \frac{2 \times 8}{3} - \frac{16}{4}-0] \\\\= 30 \pi [( \frac{16}{3} - \frac{16}{4}] \\\\= 30 \pi [( \frac{64-48}{12} - 4] \\\\= 30 \pi [( \frac{16}{12})] \\\\= 30 \pi [( \frac{4}{3})] \\\\= 30 \pi \times \frac{4}{3} \\\\= 40 \pi[/tex]
Therefore, the final volume is "[tex]40 \pi[/tex]".
Learn more about volume:
brainly.com/question/7595946
