How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = x$ $y=x$ , $x = 0$ $x=0$ , and $y = (x^{2}) - 6$ $y=(x^2)-6$ rotated around the $y = 3$ $y=3$ ?

Question

7765 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-02T18:55:22

$π \int_{0}^{3} {(3 - (x^{2} - 6))}^{2} - {(3 - x)}^{2} d x = \frac{603 π}{5}$ $piint_0^3(3-(x^2-6))^2-(3-x)^2dx=(603pi)/5$

enter image source here

The grey region is what we will be rotating around the horizontal line $y = 3$ $y=3$ .

The outer radius is $3 - (x^{2} - 6)$ $3-(x^2-6)$

The inner radius is $3 - x$ $3-x$

Using the method of washers

$π \int_{0}^{3} {(3 - (x^{2} - 6))}^{2} - {(3 - x^{2})}^{2} d x$ $piint_0^3(3-(x^2-6))^2-(3-x^2)^2dx$

$π \int_{0}^{3} {(9 - x^{2})}^{2} - {(3 - x)}^{2} d x$ $piint_0^3(9-x^2)^2-(3-x)^2dx$

$π \int_{0}^{3} 81 - 18 x^{2} + x^{4} - (9 - 6 x + x^{2}) d x$ $piint_0^3 81-18x^2+x^4-(9-6x+x^2)dx$

$π \int_{0}^{3} 81 - 18 x^{2} + x^{4} - 9 + 6 x - x^{2} d x$ $piint_0^3 81-18x^2+x^4-9+6x-x^2dx$

$π \int_{0}^{3} 72 - 19 x^{2} + x^{4} + 6 x d x$ $piint_0^3 72-19x^2+x^4+6xdx$

Integrating

$π [72 x - \frac{19}{3} x^{3} + \frac{x^{5}}{5} + 3 x^{2}]$ $pi[72x-19/3x^3+x^5/5+3x^2]$

$π [72 (3) - \frac{19}{3} {(3)}^{3} + \frac{3^{5}}{5} + 3 {(3)}^{2}]$ $pi[72(3)-19/3(3)^3+3^5/5+3(3)^2]$

$π [216 - 171 + \frac{243}{5} + 27]$ $pi[216-171+243/5+27]$

$π [72 + \frac{243}{5}]$ $pi[72+243/5]$

$π [\frac{360}{5} + \frac{243}{5}] = \frac{603 π}{5}$ $pi[360/5+243/5]=(603pi)/5$

How do you find the volume of the solid obtained by rotating the region bounded by the curves y=xy=x, x=0x=0, and y=(x2)−6y=(x^2)-6 rotated around the y=3y=3?