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Question 1167905: The region R in the first quadrant is bounded by y=6sin(πx/2), y=6(x-2)^2 and y=4x+2, and contains the point (2,6)
The volume of the solid generated by rotating R about the x-axis is:
The volume of the solid generated by rotating R about the y-axis is:
The volume of the solid with R as its base and square cross-sections perpendicular to the x-axis (and to the xy-plane) is:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! The region R is bounded by three curves in the first quadrant:
1. $y = f(x) = 6\sin(\pi x / 2)$
2. $y = g(x) = 6(x-2)^2$
3. $y = h(x) = 4x + 2$
First, let's find the intersection points of these curves to define the vertices of region R:
* **Intersection of $y=6\sin(\pi x/2)$ and $y=4x+2$:**
At $x=1$, $y = 6\sin(\pi/2) = 6$. And $y = 4(1)+2 = 6$. So, $(1,6)$ is an intersection point.
* **Intersection of $y=6(x-2)^2$ and $y=4x+2$:**
$6(x-2)^2 = 4x+2$
$6(x^2 - 4x + 4) = 4x+2$
$6x^2 - 24x + 24 = 4x+2$
$6x^2 - 28x + 22 = 0$
$3x^2 - 14x + 11 = 0$
Using the quadratic formula, $x = \frac{14 \pm \sqrt{(-14)^2 - 4(3)(11)}}{2(3)} = \frac{14 \pm \sqrt{196 - 132}}{6} = \frac{14 \pm \sqrt{64}}{6} = \frac{14 \pm 8}{6}$.
This gives $x=1$ (where $y=6$) and $x=11/3$.
For $x=11/3$, $y = 4(11/3)+2 = 44/3 + 6/3 = 50/3$. So, $(11/3, 50/3)$ is an intersection point.
* **Intersection of $y=6\sin(\pi x/2)$ and $y=6(x-2)^2$:**
At $x=1$, $y = 6\sin(\pi/2) = 6$ and $y = 6(1-2)^2 = 6$. So, $(1,6)$ is an intersection point.
At $x=2$, $y = 6\sin(\pi) = 0$ and $y = 6(2-2)^2 = 0$. So, $(2,0)$ is an intersection point.
The three vertices of the region R are $(1,6)$, $(2,0)$, and $(11/3, 50/3)$. The region contains the point $(2,6)$.
Based on these vertices and the condition that the point $(2,6)$ is contained within the region, the boundaries of the region R are defined as:
* **Upper boundary ($y_U(x)$):** $y = 4x+2$ (from $x=1$ to $x=11/3$).
* **Lower boundary ($y_L(x)$):** This is piecewise.
* From $x=1$ to $x=2$: $y = 6\sin(\pi x/2)$ (connects $(1,6)$ to $(2,0)$).
* From $x=2$ to $x=11/3$: $y = 6(x-2)^2$ (connects $(2,0)$ to $(11/3, 50/3)$).
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### 1. Volume of the solid generated by rotating R about the x-axis (Washer Method)
The volume $V_x$ is given by $V_x = \pi \int_{x_1}^{x_2} (y_{U}(x)^2 - y_{L}(x)^2) dx$.
We split the integral into two parts based on the piecewise lower boundary:
$V_x = \pi \left[ \int_{1}^{2} ((4x+2)^2 - (6\sin(\pi x/2))^2) dx + \int_{2}^{11/3} ((4x+2)^2 - (6(x-2)^2)^2) dx \right]$
**Part 1: $\int_{1}^{2} ((4x+2)^2 - 36\sin^2(\pi x/2)) dx$**
$(4x+2)^2 = 16x^2 + 16x + 4$
$36\sin^2(\pi x/2) = 36 \frac{1-\cos(\pi x)}{2} = 18 - 18\cos(\pi x)$
Integral 1 = $\int_{1}^{2} (16x^2 + 16x + 4 - (18 - 18\cos(\pi x))) dx$
$= \int_{1}^{2} (16x^2 + 16x - 14 + 18\cos(\pi x)) dx$
$= \left[ \frac{16x^3}{3} + 8x^2 - 14x + \frac{18\sin(\pi x)}{\pi} \right]_{1}^{2}$
$= \left( \frac{16(2)^3}{3} + 8(2)^2 - 14(2) + \frac{18\sin(2\pi)}{\pi} \right) - \left( \frac{16(1)^3}{3} + 8(1)^2 - 14(1) + \frac{18\sin(\pi)}{\pi} \right)$
$= \left( \frac{128}{3} + 32 - 28 + 0 \right) - \left( \frac{16}{3} + 8 - 14 + 0 \right)$
$= \left( \frac{128}{3} + 4 \right) - \left( \frac{16}{3} - 6 \right)$
$= \frac{140}{3} - \left(-\frac{2}{3}\right) = \frac{142}{3}$
**Part 2: $\int_{2}^{11/3} ((4x+2)^2 - (6(x-2)^2)^2) dx$**
$(6(x-2)^2)^2 = 36(x-2)^4$
Integral 2 = $\int_{2}^{11/3} (16x^2 + 16x + 4 - 36(x-2)^4) dx$
$= \left[ \frac{16x^3}{3} + 8x^2 + 4x - \frac{36(x-2)^5}{5} \right]_{2}^{11/3}$
$= \left( \frac{16(11/3)^3}{3} + 8(11/3)^2 + 4(11/3) - \frac{36}{5}(\frac{11}{3}-2)^5 \right) - \left( \frac{16(2)^3}{3} + 8(2)^2 + 4(2) - \frac{36}{5}(2-2)^5 \right)$
$= \left( \frac{16 \cdot 1331}{81} + \frac{8 \cdot 121}{9} + \frac{44}{3} - \frac{36}{5}(\frac{5}{3})^5 \right) - \left( \frac{128}{3} + 32 + 8 - 0 \right)$
$= \left( \frac{21296}{81} + \frac{968}{9} + \frac{44}{3} - \frac{36}{5} \frac{3125}{243} \right) - \left( \frac{128}{3} + 40 \right)$
$= \left( \frac{21296}{81} + \frac{8712}{81} + \frac{1188}{81} - \frac{7500}{81} \right) - \left( \frac{128}{3} + \frac{120}{3} \right)$
$= \frac{23696}{81} - \frac{248}{3} = \frac{23696}{81} - \frac{6696}{81} = \frac{17000}{81}$
Total Volume $V_x = \pi \left( \frac{142}{3} + \frac{17000}{81} \right) = \pi \left( \frac{142 \cdot 27}{81} + \frac{17000}{81} \right)$
$V_x = \pi \left( \frac{3834}{81} + \frac{17000}{81} \right) = \frac{20834\pi}{81}$
The volume of the solid generated by rotating R about the x-axis is: $\boxed{\frac{20834\pi}{81}}$
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### 2. Volume of the solid generated by rotating R about the y-axis (Shell Method)
The volume $V_y$ is given by $V_y = 2\pi \int_{x_1}^{x_2} x (y_{U}(x) - y_{L}(x)) dx$.
$V_y = 2\pi \left[ \int_{1}^{2} x((4x+2) - 6\sin(\pi x/2)) dx + \int_{2}^{11/3} x((4x+2) - 6(x-2)^2) dx \right]$
**Part 1: $\int_{1}^{2} x(4x+2 - 6\sin(\pi x/2)) dx = \int_{1}^{2} (4x^2 + 2x - 6x\sin(\pi x/2)) dx$**
$\int (4x^2 + 2x) dx = \frac{4x^3}{3} + x^2$
$\int -6x\sin(\pi x/2) dx$. Use integration by parts for $\int x\sin(ax)dx = -\frac{x\cos(ax)}{a} + \frac{\sin(ax)}{a^2}$. Here $a=\pi/2$.
So, $-6 \left[ -\frac{2x}{\pi}\cos(\frac{\pi x}{2}) + \frac{4}{\pi^2}\sin(\frac{\pi x}{2}) \right]_{1}^{2}$
$= -6 \left[ \left(-\frac{4}{\pi}\cos(\pi) + \frac{4}{\pi^2}\sin(\pi) \right) - \left(-\frac{2}{\pi}\cos(\frac{\pi}{2}) + \frac{4}{\pi^2}\sin(\frac{\pi}{2}) \right) \right]$
$= -6 \left[ \left(\frac{4}{\pi} + 0 \right) - \left(0 + \frac{4}{\pi^2} \right) \right]$
$= -6 \left( \frac{4}{\pi} - \frac{4}{\pi^2} \right) = -\frac{24}{\pi} + \frac{24}{\pi^2}$
For the polynomial part: $\left[ \frac{4x^3}{3} + x^2 \right]_{1}^{2} = \left(\frac{4(2)^3}{3} + (2)^2 \right) - \left(\frac{4(1)^3}{3} + (1)^2 \right) = \left(\frac{32}{3} + 4 \right) - \left(\frac{4}{3} + 1 \right) = \frac{44}{3} - \frac{7}{3} = \frac{37}{3}$
Integral 1 Value = $\frac{37}{3} - \frac{24}{\pi} + \frac{24}{\pi^2}$
**Part 2: $\int_{2}^{11/3} x((4x+2) - 6(x-2)^2) dx$**
$= \int_{2}^{11/3} x(4x+2 - 6(x^2-4x+4)) dx$
$= \int_{2}^{11/3} x(4x+2 - 6x^2+24x-24) dx$
$= \int_{2}^{11/3} (-6x^3 + 28x^2 - 22x) dx$
$= \left[ -\frac{6x^4}{4} + \frac{28x^3}{3} - \frac{22x^2}{2} \right]_{2}^{11/3}$
$= \left[ -\frac{3x^4}{2} + \frac{28x^3}{3} - 11x^2 \right]_{2}^{11/3}$
$= \left( -\frac{3(11/3)^4}{2} + \frac{28(11/3)^3}{3} - 11(11/3)^2 \right) - \left( -\frac{3(2)^4}{2} + \frac{28(2)^3}{3} - 11(2)^2 \right)$
$= \left( -\frac{3 \cdot 14641}{2 \cdot 81} + \frac{28 \cdot 1331}{3 \cdot 27} - \frac{11 \cdot 121}{9} \right) - \left( -\frac{3 \cdot 16}{2} + \frac{28 \cdot 8}{3} - 11 \cdot 4 \right)$
$= \left( -\frac{14641}{54} + \frac{37268}{81} - \frac{1331}{9} \right) - \left( -24 + \frac{224}{3} - 44 \right)$
$= \left( \frac{-43923 + 74536 - 23958}{162} \right) - \left( \frac{-72 + 224 - 132}{3} \right)$
$= \frac{6655}{162} - \frac{20}{3} = \frac{6655}{162} - \frac{1080}{162} = \frac{5575}{162}$
Total Volume $V_y = 2\pi \left( \left( \frac{37}{3} - \frac{24}{\pi} + \frac{24}{\pi^2} \right) + \frac{5575}{162} \right)$
$V_y = 2\pi \left( \frac{1998}{162} + \frac{5575}{162} - \frac{24}{\pi} + \frac{24}{\pi^2} \right)$
$V_y = 2\pi \left( \frac{7573}{162} - \frac{24}{\pi} + \frac{24}{\pi^2} \right)$
The volume of the solid generated by rotating R about the y-axis is: $\boxed{2\pi \left( \frac{7573}{162} - \frac{24}{\pi} + \frac{24}{\pi^2} \right)}$
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### 3. Volume of the solid with R as its base and square cross-sections perpendicular to the x-axis (and to the xy-plane)
The volume $V_s$ is given by $V_s = \int_{x_1}^{x_2} (y_{U}(x) - y_{L}(x))^2 dx$.
This is the same integral as for $V_x$, but without the $\pi$ factor.
$V_s = \left[ \int_{1}^{2} ((4x+2)^2 - (6\sin(\pi x/2))^2) dx + \int_{2}^{11/3} ((4x+2)^2 - (6(x-2)^2)^2) dx \right]$
Using the results from the $V_x$ calculation:
$V_s = \frac{142}{3} + \frac{17000}{81} = \frac{3834}{81} + \frac{17000}{81} = \frac{20834}{81}$
The volume of the solid with R as its base and square cross-sections perpendicular to the x-axis (and to the xy-plane) is: $\boxed{\frac{20834}{81}}$
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