How-can-I-calculate-the-volume-of-a-region-bounded-by-y-x-2-3-x-1-and-x-2-rotating-about-the-y-7-using-the-shell-method-

Question Number 131364 by EDWIN88 last updated on 04/Feb/21

How can I calculate the volume

of a region bounded by y = x^{2} + 3; x = 1

and x = 2 rotating about the y = 7 using

the shell method .

Answered by bramlexs22 last updated on 04/Feb/21

V = 2 π \underset{4}{\int^{7}} (7 - y) (\sqrt{y - 3} - 2) d y

l e t \sqrt{y - 3} = q \to \begin{array}{cc} y = 7 \to q = 2 \\ y = 4 \to q = 1 \end{array} a n d d y = 2 q d q

V = 2 π \underset{1}{\int^{2}} (7 - (q^{2} + 3)) (q - 2) (2 q d q)

V = 4 π \underset{1}{\int^{2}} (4 - q^{2}) (q^{2} - 2 q) d q

V = 4 π \underset{1}{\int^{2}} (- 8 q + 4 q^{2} + 2 q^{3} - q^{4}) d q

V = 4 π {[- 4 q^{2} + \frac{4 q^{3}}{3} + \frac{q^{4}}{2} - \frac{q^{5}}{5}]}_{1}^{2}

V = 4 π [- 4 (3) + \frac{4 (7)}{3} + \frac{15}{2} - \frac{31}{5}]

V = 4 π [\frac{- 180 + 420 - 93}{15} + \frac{15}{2}] = 4 π [\frac{147}{15} + \frac{15}{2}]

V = 4 π [\frac{298 + 225}{30}] = \frac{1046 π}{15}