Answer to Question #319578 in Calculus for Ibet

Question

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

Answer 1

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

Point of interception

$2^2=x^3$ ; $x=4^{1/3}$

Let's move to new coordinates:

$u=y-2\\t=x$

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

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

$u=t^{3/2}-2$

The required volume can be obtained by subtracting volume A

$A=\int_0^{4^{1/3}} \pi(t^{3/2}-2)^2dt$

from volume B

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

$V=B-A=\int_0^{4^{1/3}} \pi(2^2-(t^{3/2}-2)^2)dt=$$\int_0^{4^{1/3}} \pi(-t^3+4t^{3/2})dt=\pi(-\frac{t^4}{4}+\frac85t^{5/2})|_0^{4^{1/3}}=$$\pi(-\frac{4^{4/3}}{4}+\frac854^{5/6})=\frac{\pi \cdot4^{1/3}\cdot 11}{5}\approx10.97$