Question #319578

Determine the volume of the solid generated when the region bounded by the curve y^2=x^3 and the x-axis, is revolved about the line y=2

1
Expert's answer
2022-04-01T05:12:14-0400

Region bounded by the curve y2=x3y^2=x^3 and the x-axis, is revolved about the line y=2y=2.

Point of interception

22=x32^2=x^3 ; x=41/3x=4^{1/3}



Let's move to new coordinates:

u=y2t=xu=y-2\\t=x

Now region is revolved about the line u=0u=0

(u+2)2=t3(u+2)^2=t^3

u=t3/22u=t^{3/2}-2



The required volume can be obtained by subtracting volume A


A=041/3π(t3/22)2dtA=\int_0^{4^{1/3}} \pi(t^{3/2}-2)^2dt

from volume B


B=041/3π(2)2dt=041/3π22dtB=\int_0^{4^{1/3}} \pi(-2)^2dt=\int_0^{4^{1/3}} \pi2^2dt .

V=BA=041/3π(22(t3/22)2)dt=V=B-A=\int_0^{4^{1/3}} \pi(2^2-(t^{3/2}-2)^2)dt=041/3π(t3+4t3/2)dt=π(t44+85t5/2)041/3=\int_0^{4^{1/3}} \pi(-t^3+4t^{3/2})dt=\pi(-\frac{t^4}{4}+\frac85t^{5/2})|_0^{4^{1/3}}=π(44/34+8545/6)=π41/311510.97\pi(-\frac{4^{4/3}}{4}+\frac854^{5/6})=\frac{\pi \cdot4^{1/3}\cdot 11}{5}\approx10.97


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS