smc-1180-65-Surface area

A parametric curve is given by \(x=k t, y=e^{k t}\), where \(k\) is a positive constant. The curve is rotated about the \(x\)-axis from \(t=a\) to \(t=b\), where \(b>a\), to form a surface of revolution.

The area of this surface is given by

\(2 \displaystyle \pi \int_a^b e^{k t} \sqrt{k^2 t^2+e^{2 k t}}\ d t\)
\(2 \displaystyle\pi \sqrt{k} \int_a^b e^{k t} \sqrt{1+e^{k t}}\ d t\)
\(2 \displaystyle\pi \int_{\Large{e^{k a}}}^{\Large{e^{k b}}} \sqrt{1+u^2}\ d u\)
\(2 \displaystyle\pi \int_{\Large{e^{k a}}}^{\Large{e^{k b}}} u \sqrt{1+u^2}\ d u\)

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\(C\)

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\(\text{Using formula for S.A. of a curve rotated about \(x\)-axis:}\)

\(\text{S.A. }\)	\(=2 \pi \displaystyle \int_a^b x \sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2} \, dt\)
	\(=2 \pi \displaystyle\int_a^b e^{k t} \sqrt{k^2+k^2 e^{2 k t}} \, dt\)

♦ Mean mark 49%.

\(\text{Using substitution} \ \ u=e^{kt}:\)

\(\text{S.A.}=2 \pi \displaystyle\int_{e^{ka}}^{e^{k b}} \frac{1}{k} \sqrt{k^2\left(1+u^2\right)} \, du\)

\(\Rightarrow C\)

The curve given by \(y^2=x-1\), where \(2 \leq x \leq 5\), is rotated about the \(x\)-axis to form a solid of revolution.

i. Write down the definite integral, in terms of \(x\), for the volume of this solid of revolution. (1 mark)

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ii. Find the volume of the solid of revolution. (1 mark)

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i. Express the curved surface area of the solid in the form \(\pi \displaystyle \int_a^b \sqrt{A x-B}\, d x\), where \(a, b, A, B\) are all positive integers. (2 marks)

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ii. Hence or otherwise, find the curved surface area of the solid correct to three decimal places. (1 mark)

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The total surface area of the solid consists of the curved surface area plus the areas of the two circular discs at each end.

The 'efficiency ratio' of a body is defined as its total surface area divided by the enclosed volume.

Find the efficiency ratio of the solid of revolution correct to two decimal places. (2 marks)

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Another solid of revolution is formed by rotating the curve given by \(y^2=x-1\) about the \(x\)-axis for \(2 \leq x \leq k\), where \(k \in R\). This solid has a volume of \(24 \pi\).
Find the efficiency ratio for this solid, giving your answer correct to two decimal places. (3 marks)

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a.i. \(\displaystyle V=\pi \int y^2 d x=\pi \int_2^5(x-1) dx\)

a.ii. \(\text{Evaluating integral (by calc):}\)

\(V=\dfrac{15 \pi}{2}\ \text{u}^3\)

b.i. \(\displaystyle y=\sqrt{x-1} \ \Rightarrow \ \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{x-1}} \ \Rightarrow \ \frac{d^2 y}{dx^2}=\frac{1}{4(x-1)}\)

\(\ \ \begin{aligned} \displaystyle \int_2^5 2 \pi \sqrt{x-1} \ \sqrt{1+\frac{1}{4(x-1)}} \ d x & =\int_2^5 2 \pi \sqrt{x-1+\frac{1}{4}}\, d x \\ & =\pi \int_2^5 \sqrt{4 x-3}\, d x\end{aligned}\)

b.ii. \(\text{Evaluate integral in b.i.}\)

\(\text {S.A.}=30.847\ \text{u}^2\)

c. \(y=\sqrt{x-1}\)

\(\text{At}\ \ x=2\ \ \Rightarrow y=1\)

\(\text{At}\ \ x=5\ \ \Rightarrow y=2\)

\begin{aligned}
\text{Total S.A. } &=30.847+\pi(1)^2+\pi(2)^2 \\
&=46.5545\ \text{u}^2 \\
\end{aligned}

\(\text {Efficiency ratio}=\dfrac{46.5545}{\frac{15 \pi}{2}}=1.98\ \text{(2 d.p.)}\)

d. \(V=\displaystyle \pi \int_2^k(x-1) d x=24 \pi\)

\(\text{Solve (by calc):}\)

\(\dfrac{\pi k(k-2)}{2}=24 \pi \ \Rightarrow \ k=8 \)

\(\text {S.A.}=\pi \displaystyle{\int_2^8} \sqrt{4 x-3}\, d x=75.9163 \ldots\)

\(\text{Total S.A.}=75.9163+\pi(1+7)=101.049 \)

\(\text{Efficiency ratio}=\dfrac{101.049}{24 \pi}= 1.34\ \text{(2 d.p.)}\)