Question 5
Figure 2 shows the graph of part of the curve $C$ with equation $y=\sqrt{2x+6}$. The finite region enclosed by the curve $C$ and the straight line with equation $3y-x=3$ is rotated through $360^\circ$ about the $x$-axis.
Use algebraic integration to find the exact volume of the solid generated.
Give your answer in terms of $\pi$
(8)
Solution:
We are given the curve $C$ with equation $y = \sqrt{2x + 6}$ and the straight line with equation $3y - x = 3$. The finite region enclosed by these two curves is rotated through $360^\circ$ about the $x$-axis. We are tasked with finding the exact volume of the solid generated by this rotation using algebraic integration.
Step 1: Express the equation of the line in terms of $y$
The given straight line equation is:
$$ 3y - x = 3 $$Solving for $y$:
$$ 3y = x + 3 $$ $$ y = \frac{x + 3}{3} $$Thus, the equation of the line is $y = \frac{x + 3}{3}$.
Step 2: Find the points of intersection between the curve and the line
To find the points of intersection, we set the expressions for $y$ from the curve and the line equal to each other:
$$ \sqrt{2x + 6} = \frac{x + 3}{3} $$Now, square both sides to eliminate the square root:
$$ 2x + 6 = \left( \frac{x + 3}{3} \right)^2 $$ $$ 2x + 6 = \frac{(x + 3)^2}{9} $$Multiply both sides by 9 to clear the denominator:
$$ 9(2x + 6) = (x + 3)^2 $$ $$ 18x + 54 = x^2 + 6x + 9 $$Rearrange the equation:
$$ x^2 + 6x + 9 - 18x - 54 = 0 $$ $$ x^2 - 12x - 45 = 0 $$Now, solve this quadratic equation using the quadratic formula:
$$ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(-45)}}{2(1)} $$ $$ x = \frac{12 \pm \sqrt{144 + 180}}{2} $$ $$ x = \frac{12 \pm \sqrt{324}}{2} $$ $$ x = \frac{12 \pm 18}{2} $$Thus, the solutions for $x$ are:
$$ x = \frac{12 + 18}{2} = 15 \quad \text{and} \quad x = \frac{12 - 18}{2} = -3 $$So, the points of intersection occur at $x = -3$ and $x = 15$.
Step 3: Set up the volume integral
The volume of the solid generated by rotating a region around the $x$-axis is given by the formula:
$$ V = \pi \int_{a}^{b} \left( f(x)^2 - g(x)^2 \right) \, dx $$where $f(x)$ and $g(x)$ represent the functions describing the outer and inner radii of the solid, respectively. In this case:
- $f(x) = \sqrt{2x + 6}$ is the curve (outer radius),
- $g(x) = \frac{x + 3}{3}$ is the line (inner radius).
Thus, the volume integral becomes:
$$ V = \pi \int_{-3}^{15} \left( \left( \sqrt{2x + 6} \right)^2 - \left( \frac{x + 3}{3} \right)^2 \right) \, dx $$ $$ V = \pi \int_{-3}^{15} \left( (2x + 6) - \frac{(x + 3)^2}{9} \right) \, dx $$Step 4: Simplify the integrand
First, expand $\frac{(x + 3)^2}{9}$:
$$ \frac{(x + 3)^2}{9} = \frac{x^2 + 6x + 9}{9} $$So, the integrand becomes:
$$ 2x + 6 - \frac{x^2 + 6x + 9}{9} $$To simplify, rewrite the expression with a common denominator:
$$ 2x + 6 = \frac{18x + 54}{9} $$Thus, the integrand is:
$$ \frac{18x + 54}{9} - \frac{x^2 + 6x + 9}{9} = \frac{18x + 54 - x^2 - 6x - 9}{9} $$ $$ = \frac{-x^2 + 12x + 45}{9} $$The volume integral is now:
$$ V = \pi \int_{-3}^{15} \frac{-x^2 + 12x + 45}{9} \, dx $$Step 5: Integrate the expression
We can factor out $\frac{1}{9}$ from the integral:
$$ V = \frac{\pi}{9} \int_{-3}^{15} (-x^2 + 12x + 45) \, dx $$Now, integrate term by term:
$$ \int -x^2 \, dx = -\frac{x^3}{3}, \quad \int 12x \, dx = 6x^2, \quad \int 45 \, dx = 45x $$Thus, the integral becomes:
$$ V = \frac{\pi}{9} \left[ -\frac{x^3}{3} + 6x^2 + 45x \right]_{-3}^{15} $$Step 6: Evaluate the definite integral
First, evaluate the expression at $x = 15$:
$$ -\frac{15^3}{3} + 6(15^2) + 45(15) = -\frac{3375}{3} + 6(225) + 675 = -1125 + 1350 + 675 = 900 $$Next, evaluate the expression at $x = -3$:
$$ -\frac{(-3)^3}{3} + 6(-3)^2 + 45(-3) = -\frac{-27}{3} + 6(9) - 135 = 9 + 54 - 135 = -72 $$Thus, the definite integral is:
$$ 900 - (-72) = 900 + 72 = 972 $$Step 7: Final volume calculation
Now, multiply by $\frac{\pi}{9}$:
$$ V = \frac{\pi}{9} \times 972 = 108\pi $$Thus, the exact volume of the solid generated is:
$$ \boxed{108\pi} $$Question 6
Figure 3 shows a right pyramid with a horizontal square base.
$$ \begin{aligned} &AB = BC = CD = DA = x \mathrm{~cm} \\ &AV = BV = CV = DV = x \mathrm{~cm} \end{aligned} $$
$O$ is the point of intersection of the diagonals of the base.
The vertex $V$ of the pyramid is vertically above $O$
(a) Show that $VO=\frac{\sqrt{2}}{2}x \mathrm{~cm}$ (3)
(b) Find, in degrees, the size of the angle $AVC$ (2)
(c) Find, in degrees to one decimal place, the size of the angle between the plane $VAB$ and the plane $VDC$ (3)
The volume of the pyramid is $200 \mathrm{~cm}^3$
Given that the volume of a pyramid $=\frac{1}{3}\times$ base area $\times$ height
(d) Find to 3 significant figures, the value of $x$ (3)
Solution
Figure 3 shows a right pyramid with a horizontal square base. The following are given:
$$ AB = BC = CD = DA = x \text{ cm}, $$ $$ AV = BV = CV = DV = x \text{ cm}. $$$O$ is the point of intersection of the diagonals of the square base, and the vertex $V$ is vertically above $O$.
(a) Show that $VO = \frac{\sqrt{2}}{2}x$
The diagonals of the square base intersect at $O$, dividing each diagonal into two equal parts. Let $AC$ and $BD$ be the diagonals.
The length of a diagonal of the square can be found using the Pythagorean theorem:
$$ AC = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2}. $$Since $O$ is the midpoint of $AC$, we have:
$$ AO = \frac{AC}{2} = \frac{x\sqrt{2}}{2}. $$The vertex $V$ is vertically above $O$, so $VO$ forms the height of the right triangle $AVO$, where $AV = x$ and $AO = \frac{x\sqrt{2}}{2}$. Using the Pythagorean theorem:
$$ AV^2 = AO^2 + VO^2 $$ $$ x^2 = \left(\frac{x\sqrt{2}}{2}\right)^2 + VO^2 $$ $$ x^2 = \frac{2x^2}{4} + VO^2 $$ $$ VO^2 = x^2 - \frac{x^2}{2} = \frac{x^2}{2} $$ $$ VO = \sqrt{\frac{x^2}{2}} = \frac{x\sqrt{2}}{2}. $$Thus, $VO = \frac{\sqrt{2}}{2}x$ cm.
(b) Find the size of $\angle AVC$
Consider the triangle $AVC$. The sides of the triangle are:
$$ AV = x, \quad VC = x, \quad AC = x\sqrt{2}. $$Using the cosine rule:
$$ \cos \angle AVC = \frac{AV^2 + VC^2 - AC^2}{2 \cdot AV \cdot VC} $$ $$ \cos \angle AVC = \frac{x^2 + x^2 - (x\sqrt{2})^2}{2 \cdot x \cdot x} $$ $$ \cos \angle AVC = \frac{x^2 + x^2 - 2x^2}{2x^2} $$ $$ \cos \angle AVC = \frac{0}{2x^2} = 0. $$Since $\cos \angle AVC = 0$, $\angle AVC = 90^\circ$.
(c) Find the angle between the planes $VAB$ and $VDC$
Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The line $MN$ is parallel to the square base and lies in the plane $ABCD$. The angle between the planes $VAB$ and $VDC$ is equal to the angle $\angle MVN$.
From the geometry of the pyramid:
- $VM = VN = \sqrt{VO^2 + OM^2}$, where $OM = \frac{AB}{2} = \frac{x}{2}$.
- $VO = \frac{x\sqrt{2}}{2}$ (from part (a)).
Using the Pythagorean theorem in $\triangle VOM$:
$$ VM = \sqrt{VO^2 + OM^2} = \sqrt{\left(\frac{x\sqrt{2}}{2}\right)^2 + \left(\frac{x}{2}\right)^2} $$ $$ VM = \sqrt{\frac{x^2}{2} + \frac{x^2}{4}} = \sqrt{\frac{2x^2 + x^2}{4}} = \sqrt{\frac{3x^2}{4}} = \frac{x\sqrt{3}}{2}. $$The length $MN = AB = x$. The angle $\angle MVN$ can be found using the cosine rule:
$$ \cos \angle MVN = \frac{VM^2 + VN^2 - MN^2}{2 \cdot VM \cdot VN} $$ $$ \cos \angle MVN = \frac{\left(\frac{x\sqrt{3}}{2}\right)^2 + \left(\frac{x\sqrt{3}}{2}\right)^2 - x^2}{2 \cdot \frac{x\sqrt{3}}{2} \cdot \frac{x\sqrt{3}}{2}} $$ $$ \cos \angle MVN = \frac{\frac{3x^2}{4} + \frac{3x^2}{4} - x^2}{\frac{3x^2}{2}} $$ $$ \cos \angle MVN = \frac{\frac{6x^2}{4} - x^2}{\frac{3x^2}{2}} = \frac{\frac{2x^2}{4}}{\frac{3x^2}{2}} = \frac{1}{3}. $$Thus:
$$ \angle MVN = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ. $$(d) Find the value of $x$
The volume of the pyramid is given as $200$ cm$^3$. The formula for the volume of a pyramid is:
$$ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}. $$The base area is the area of the square:
$$ \text{Base Area} = x^2. $$The height is $VO = \frac{x\sqrt{2}}{2}$ (from part (a)). Substituting these values:
$$ 200 = \frac{1}{3} \times x^2 \times \frac{x\sqrt{2}}{2} $$ $$ 200 = \frac{x^3\sqrt{2}}{6} $$ $$ x^3 = \frac{200 \cdot 6}{\sqrt{2}} = \frac{1200}{\sqrt{2}} = 600\sqrt{2} $$ $$ x = \sqrt[3]{600\sqrt{2}} = 9.467 \approx 9.47 \text{ cm} \, \text{(to 3 significant figures)}. $$
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