Compute the volume generated by rotating the region bounded by the graphs of the given equations:
First, sketch all of the given functions and identify a common area among them:
Finally, choose the proper method of volumetric intergation. Notice how the differencial line is parallel to the axis of rotation. Because of this, use the shell method: Let \(R = x + 1\) and \(H = (x \:- 2)^3 \:- (x \:- 2)= x^3 \:- 6x^2 + 11x \:- 6 \)
\(
\begin{align*}
& V = 2\pi \int^2_1 [(x+1)(x^3 \:- 6x^2 + 11x \:- 6)]dx = 2\pi \int^1_0 [x^4 \:- 5x^3 + 5x^2 + 5x \:- 6]dx \\[5pt]
& V = 2\pi \left[ \frac{1}{5}x^5 \:- \frac{5}{4}x^4 + \frac{5}{3}x^3 + \frac{5}{2}x^2 \:- 6x \right] \Bigg\rvert^2_1 \\[5pt]
& V = 2\pi \left[ \left( \frac{1}{5}(2)^5 \:- \frac{5}{4}(2)^4 + \frac{5}{3}(2)^3 + \frac{5}{2}(2)^2 \:- 6(2) \right) \:- \left( \frac{1}{5}(1)^5 \:- \frac{5}{4}(1)^4 + \frac{5}{3}(1)^3 + \frac{5}{2}(1)^2 \:- 6(1)\right) \right] \\[5pt]
& V = 2\pi \left[ \left( \frac{32}{5} \:- 20 + \frac{40}{3} + 10 \:- 12 \right) \:- \left( \frac{1}{5} \:- \frac{5}{4} + \frac{5}{3} + \frac{5}{2} \:- 6 \right) \right] = 2\pi \left[-\frac{34}{15} -\left(-\frac{173}{60}\right)\right] \\[5pt]
& \color{green}{V = \frac{37}{30}\pi}
\end{align*}
\)
First, sketch all of the given functions and identify a common area among them:
The problem remains mostly unchanged, the only modification being the radius \(R = 2 \:- x\). Updating the previous integral with this change yields:
\(
\begin{align*}
& V = 2\pi \int^2_1 (2 \:- x)(x^3 \:- 6x^2 + 11x \:- 6) dx = 2\pi \int^1_0 (-x^4 + 8x^3 \:- 23x^2 + 28x \:- 12)dx \\[5pt]
& V = 2\pi \left[ -\frac{1}{5}x^5 + \frac{8}{4}x^4 \:- \frac{23}{3}x^3 +\frac{28}{2}x^2 \:- 12x \right] \Bigg\rvert^2_1 = 2\pi \left[ -\frac{1}{5}x^5 + 2x^4 \:- \frac{23}{3}x^3 + 14x^2 \:- 12x \right] \Bigg\rvert^2_1 \\[5pt]
& V = 2\pi \left( \left[ -\frac{1}{5}(2)^5 + 2(2)^4 \:- \frac{23}{3}(2)^3 + 14(2)^2 \:- 12(2) \right] \:- \left[ -\frac{1}{5}(1)^5 + 2(1)^4 \:- \frac{23}{3}(1)^3 + 14(1)^2 \:- 12(1) \right] \right) \\[5pt]
& V = 2\pi \left[ \left(-\frac{32}{5} + 32 \:- \frac{184}{3} + 56 \:- 24 \right) \:- \left(-\frac{1}{5} + 2 \:- \frac{23}{3} + 14 \:- 12\right) \right] = 2\pi\left[-\frac{56}{15}-\left(-\frac{58}{15}\right)\right] \\[5pt]
& \color{green}{V = \frac{4}{15}\pi}
\end{align*}
\)
First, sketch all of the given functions and identify a common area among them:
Now, as we can’t change the orientation of the differential line, change the method to washer (because of the perperndicular position between the differential and the rotation axis):
\(
\begin{align*}
& R = (x \:- 2)^3 + 1; \Rightarrow R^2 = ((x \:- 2)^3 + 1)^2 = x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 228x^2 \:- 168x + 49 \\[5pt]
& r = (x \:- 2) + 1; \Rightarrow r = (x \:- 1)^2 = x^2 \:- 2x + 1 \\[10pt]
& R^2 \:- r^2 = x^6 \:- 12x^5 + 60x^4 \:- 158x^3 + 228x^2 \:- 168x + 49 \:- ( x^2 \:- 2x + 1) \\[5pt]
& = x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 227x^2 \:- 166x+ 48 \\\\[5pt]
& V = \pi \int^2_1 \left( x^6 \:- 12x^5 + 60x^4 \:-158x^3 + 227x^2 \:- 166x+ 48 \right)dx \\[5pt]
& V = \pi\left[ \frac{1}{7}x^7 \:- \frac{12}{6}x^6 + \frac{60}{5}x^5 \:- \frac{158}{4}x^4 + \frac{227}{3}x^3 \:- \frac{166}{2}x^2 + 48x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left[ \frac{1}{7}x^7 \:- 2x^6 + 12x^5 \:- \frac{79}{2}x^4 + \frac{227}{3}x^3 \:- 83x^2 + 48x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left( \left[ \frac{1}{7}(2)^7 \:- 2(2)^6 + 12(2)^5 \:- \frac{79}{2}(2)^4 + \frac{227}{3}(2)^3 \:- 83(2)^2 + 48(2) \right] \:- \left[ \frac{1}{7}(1)^7 \:- 2(1)^6 + 12(1)^5 \:- \frac{79}{2}(1)^4 + \frac{227}{3}(1)^3 \:- 83(1)^2 + 48(1) \right] \right) \\[5pt]
& V = \pi \left[ \left( \frac{128}{7} \:- 128 + 384 \:- 632 + \frac{1816}{3} \:- 332 + 96 \right) \:- \left( \frac{1}{7} \:- 2 + 12 \:- \frac{79}{2} + \frac{227}{3} \:- 83 + 48 \right) \right] = \pi \left[ \left(\frac{244}{21}\right) \:- \left(\frac{475}{12} \right) \right] \\[5pt]
& \color{green}{V = \frac{13}{42}\pi}
\end{align*}
\)
First, sketch all of the given functions and identify a common area among them:
The problem remains mostly unchanged (we are still using the washer method), so the only modification being both radiuses:
\(
\begin{align*}
& R = 1 \:- (x \:- 2); \Rightarrow r = (3 \:- x)^2 = 9 \:- 6x + x^2 \\[10pt]
& r = 1 \:- (x \:- 2)^3; \Rightarrow R^2 = (1 \:- (x \:- 2)^3)^2 = x^6 \:- 12x^5 + 60x^4 \:- 162x^3 + 252x^2 \:- 216x + 81 \\[5pt]
& R^2 \:- r^2 = ( 9 \:- 6x + x^2) \:- (x^6 \:- 12x^5 + 60x^4 \:- 162x^3 + 252x^2 \:- 216x + 81) \\[5pt]
& = -x^6 + 12x^5 \:- 60x^4 + 162x^3 \:- 251x^2 + 210x \:- 72 \\\\[5pt]
& V = \pi \int^2_1 \left( -x^6 + 12x^5 \:- 60x^4 + 162x^3 \:- 251x^2 + 210x \:- 72 \right)dx \\[5pt]
& V = \pi\left[ -\frac{1}{7}x^7 + \frac{12}{6}x^6 \:- \frac{60}{5}x^5 + \frac{162}{4}x^4 \:- \frac{251}{3}x^3 + \frac{210}{2}x^2 \:- 72x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left[ -\frac{1}{7}x^7 + 2x^6 \:- 12x^5 + \frac{81}{2}x^4 \:- \frac{251}{3}x^3 + 105x^2 \:- 72x \right] \Bigg\rvert^2_1 \\[5pt]
& V = \pi\left( \left[ -\frac{1}{7}(2)^7 + 2(2)^6 \:- 12(2)^5 + \frac{81}{2}(2)^4 \:- \frac{251}{3}(2)^3 + 105(2)^2 \:- 72(2) \right] \:- \left[ -\frac{1}{7}(1)^7 + 2(1)^6 \:- 12(1)^5 + \frac{81}{2}(1)^4 \:- \frac{251}{3}(1)^3 + 105(1)^2 \:- 72(1) \right] \right) \\[5pt]
& V = \pi \left( \left[ -\frac{128}{7} + 128 \:- 384 + 648 \:- \frac{2008}{3} + 420 \:- 144 \right] \:- \left[ -\frac{1}{7} + 2 \:- 12 + \frac{81}{2} \:- \frac{251}{3} + 105 \:- 72 \right] \right) \\[5pt]
& V = \pi\left( \left[ -\frac{412}{21} \right] \:- \left[ -\frac{853}{42} \right] \right) \\[5pt]
& \color{green}{V = \frac{29}{42}\pi}
\end{align*}
\)