Question

1. Compute the volume of a solid using DISK/WASHER METHOD for the functions bounded by y = x³ and y = x revolve about the axis at y = -2 NO. 2. Using SHELL/CYLINDRICAL METHOD, calculate the volume of a solid bounded by y = x and y = √x revolve around the axis at x = -4.​

Answer 1

1)
`\displaystyle (25\pi)/(21)` cubic units

2)
`\displaystyle (22\pi)/(15)` cubic units

Explanation:

Problem 1

Because our axis of rotation is not the x-axis, we must use the washer method, which is
`\displaystyle A=\pi\int\limits^b_a \bigr[{\text{(Outer Radius)}^2-\text{(Inner Radius)}^2)}\bigr] \, dx`.

If we plot the graphs of each function, we see that
`y=x` is our outer function, and
`y=x^3` is our inner function. If we set the equations equal to each other, we can get our bounds of integration where the equations intersect, which are
`a=0` to
`b=1`.

Taking into consideration our axis of revolution is
`y=-2` and NOT
`y=0`, the x-axis, the outer radius would be
`-2-x`, and our inner radius would be
`-2-x^3`.

Putting everything together, we calculate the definite integral:

`\displaystyle A=\pi\int\limits^b_a \bigr[{\text{(Outer Radius)}^2-\text{(Inner Radius)}^2)}\bigr] \, dx\\\\A=\pi\int\limits^1_0 \bigr[(-2-x)^2-(-2-x^3)^2}\bigr] \, dx\\\\A=\pi\int\limits^1_0 \bigr[(x^2+4x+4)-(x^6+4x^3+4)}\bigr] \, dx\\\\A=\pi\int\limits^1_0 \bigr[x^2+4x-x^6-4x^3}\bigr] \, dx\\\\A=\pi\int\limits^1_0 \bigr[-x^6-4x^3+x^2+4x}\bigr] \, dx\\\\A=\pi\biggr[-(x^7)/(7)-x^4+(x^3)/(3)+2x^2\biggr]^1_0\\\\A=\pi\biggr[-((1)^7)/(7)-(1)^4+((1)^3)/(3)+2(1)^2\biggr]`

`\displaystyle A=\pi\biggr[-(1)/(7)-1+(1)/(3)+2\biggr]\\\\A=\pi\biggr[-(3)/(21)-(21)/(21) +(7)/(21)+(42)/(21)\biggr]\\\\A=(25\pi)/(21)\approx3.74\:\text{cubic\:units}`

Problem 2

The shell method for revolving the area between the two curves
`f(x)` and
`g(x)` about
`x=h` is
`\displaystyle A=2\pi\int\limits^b_a {(x-h)\bigr[f(x)-g(x)\bigr]} \, dx` where
`f(x)\geq g(x)`.

By graphing the two equations, we can see that
`f(x)=√(x)` is the upper function, and
`g(x)=x` is the lower function. If we set the two equations equal to each other and find the points of intersection, we can determine our bounds of integration. Since they intersect at points
`(0,0)` and
`(1,1)`, our bounds of integration will be
`a=0` to
`b=1`.

Since our axis of revolution is
`x=-4` and NOT
`x=0`, the y-axis, we must account for this as indicated in the above equation.

Putting everything together, we can calculate the definite integral:

`\displaystyle A=2\pi\int\limits^b_a {(x-h)\bigr[f(x)-g(x)\bigr]} \, dx\\\\A=2\pi\int\limits^1_0 {(x-(-4))\bigr[√(x)-x\bigr]} \, dx\\\\A=2\pi\int\limits^1_0 {(x+4)\bigr[√(x)-x\bigr]} \, dx\\\\A=2\pi\int\limits^1_0 {\bigr[x√(x)-x^2+4√(x)-4x\bigr]} \, dx\\\\A=2\pi\int\limits^1_0 {\bigr[x^(3)/(2) -x^2+4x^{(1)/(2)}-4x\bigr]} \, dx\\\\A=2\pi\biggr[(2x^(5)/(2) )/(5)-(x^3)/(3)+(8x^(3)/(2))/(3)-2x^2\biggr]^1_0`

`\displaystyle A=2\pi\biggr[(2(1)^(5)/(2) )/(5)-((1)^3)/(3)+(8(1)^(3)/(2))/(3)-2(1)^2\biggr]\\\\A=2\pi\biggr[(2)/(5)-(1)/(3)+(8)/(3)-2\biggr]\\\\A=2\pi\biggr[(2)/(5)+(7)/(3)-2\biggr]\\\\A=2\pi\biggr[(6)/(15)+(35)/(15)-(30)/(15)\biggr]\\\\A=2\pi\biggr[(11)/(15)\biggr]\\\\A=(22\pi)/(15)\approx4.61\:\text{cubic\:units}`