Question

a. Calculate the volume of the solid of revolution created by rotating the curve y = 3 + 3 exp(-5 x) about the x-axis, for x between 2 and 4. Volume : 56.55 b. The equation of a circle of radius r, centered at the origin (0,0), is given by r^2 = x^2 + y^2 Rearrange this equation to find a formula for y in terms of x and r. (Take the positive root.) What solid of revolution is swept out if this curve is rotated around the x axis, and x is allowed to vary between -r and r Suppose we wanted to set the following integral so that V gives the volume of a sphere of radius r V = integral a to b f(x) dx What would a, b and f(x) be?a=b=f(x)=Carry out the integration, and calculate the value of V in terms of r.V=

Answer 1

Answer:

A) The volume of solid is 18π cubic unit.

B) The volume of sphere is $\dfrac{4}{3}\pi r^3$

    a = -r , b = r , $f(x)=\sqrt{r^2-x^2}$ and $V=\dfrac{4}{3}\pi r^3$

Solution A)

The given curve, $y=3+3e^{-5x}$ rotate about x-axis between 2 to 4.

Please find attachment for solid figure or rotation.

Using disk method to find the volume rotation about x-axis

$V=\int_a^b\pi R^2dx$  

where, a = 2, b= 4 , $R=y=3+3e^{-5x}$ and dx is thickness of disk.

$V=\int_2^4\pi (3+3e^{-5x})^2dx$

$V=9\pi\int_2^4(1+e^{-10x}+2e^{-5x})dx$

$V=9\pi(x-\dfrac{1}{10}e^{-10x}-\dfrac{2}{5}e^{-5x})|_2^4)$

$V=9\pi(4-\dfrac{1}{10}e^{-40}-\dfrac{2}{5}e^{-20}-2+\dfrac{1}{10}e^{-20}-\dfrac{2}{5}e^{-10}))$

$V=9\pi (2-0)$

$V=18\pi$

Hence, the volume of solid is 18π cubic unit

Solution B)

The equation of circle of radius r and centered at origin (0,0).

$x^2+y^2=r^2$

$y=\sqrt{r^2-x^2}$

The solid form is sphere of radius r.

Using disk method, to find volume of solid

$V=\int_a^b\pi R^2dx$  

where, a = -r, b = r , $R=y=\sqrt{r^2-x^2}$ and dx is thickness of disk.

$V=\int_{-r}^r\pi (r^2-x^2)dx$

$V=\pi(r^2x-\dfrac{x^3}{3})|_{-r}^r$

$V=\pi(2r^3-\dfrac{2r^3}{3})$

$V=\dfrac{4}{3}\pi r^3$

Hence, the volume of sphere is $\dfrac{4}{3}\pi r^3$