Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y  1 3 ex2 /3 , y  0, x  0, and x  3 about the y-axis. Round your answer to three decimal places.

Answer 1

Explanation:

`y = f(x) =(1)/(√(3 \pi) ) e^{-x^(2/3)}`

y = 0, x = 0 and x = 3

Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by

`dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) (1)/(√(3\pi))e^{-x^{(2)/(3)}}`

`dV=2\sqrt{(\pi)/(3)}xe^{-x^{(2)/(3)}}dx`

Integrate the above integral over the limits x=0 to x=3 which implies

`\int_(0)^(V)dV=2\sqrt{(\pi)/(3)}\int_(0)^(3)xe^{-x^{(2)/(3)}}dx`

Solve by subsititution

`Let,\\ -x^{(2)/(3)}=y => (-2)/(3)x^{(-1)/(3)}dx=dy => x^{(-1)/(3)}dx=(-3)/(2)dy`

Also, apply the new limits

`At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}`

This implies,

`\int_(0)^(V)dV=2\sqrt{(\pi)/(3)}\int_(0)^(3)x^{(4)/(3)}e^{-x^{(2)/(3)}}x^{(-1)/(3)}dx=2\sqrt{(\pi)/(3)}\int_(0)^{-\sqrt[3]{9}}y^(2)e^(y)((-3)/(2))dy`

`V=-√(3\pi)\int_(0)^{-\sqrt[3]{9}}y^(2)e^(y)dy`

Let,

`I=\int_(0)^{-\sqrt[3]{9}}y^(2)e^(y)dy`

Integrate by parts the above integral

`u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y`

Integrate by parts formula

`\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy`

Again integrate by parts

`u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y`

Integrate by parts formula

`\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2]`

Therefore,

`I=[e^y(y^2-2y+2)]_(0)^{-\sqrt[3]{9}}\\\\=e^(-2.0802)[(2.0802)^2+2(2.0802)+2]-e^(0)[0-0+2]\\\\((4.3272+4.1604+2))/(8.0061)-2\\\\=(10.4876)/(8.0061)-2\\\\=1.3099-2\\\\=-0.6901`

This implies, the volume is

`V=-√(3\pi)I\\\\=-√(3* 3.142) * (-0.6901)\\\\=3.0701 * 0.6901\\\\=2.1186`

That is, up to three decimal places

`V\approx 2.118`