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Applications of integration: DHS 2024 Promo Q13

Dunman High School2024 PromoPromo P1 Q13●●● Challenging14 marks

What this question tests

Integration (Applications).

Question

. An art student wants to create an art installation “Glimpse of Hope” on the wall. The installation consists of a drawing, which is modelled by the curve y=2x2e(2−x)y = 2x^2\mathrm{e}^{(2 - x)} and the line y=2exy = 2\mathrm{e} x. The two graphs intersect at the origin and touch at the point (1,2e)(1, 2\mathrm{e}). The student intends to fit lightbulbs into the region RR to signify that there is always hope in every “up-and-down” situation in life.

Dunman High School 2024 Promo Q13: question diagram

  1. Find the exact area of region RR. [4]

  2. Dunman High School 2024 Promo Q13: question diagram

    The light bulbs that the student used are specially manufactured for the art installation. Each light bulb consists of part of an empty sphere with radius 66 cm and a solid cylinder of radius 252\sqrt{5} cm and height 33 cm as shown in the diagram. The bottom of the sphere is sliced off so that its edges touches the edges of the top of the cylinder. An inert gas is used to fill the spherical part of the light bulb to prevent the filament from decaying.

    The sphere of the light bulb is modelled by rotating part of the curve x2+(y−6)2=36x^2 + (y - 6)^2 = 36 around the yy-axis as shown in the diagram above.

    1. Assuming that the filament takes up negligible space and that the light bulb is of negligible thickness, calculate the exact volume of the spherical part of the light bulb that needs to be filled with the inert gas. [4]

    2. To determine the amount of special coating to apply to the exterior of the spherical part of the light bulb, the manufacturer wishes to find the surface area of the spherical part of the light bulb. It is given that the area of a surface of revolution formed by revolving the graph of x=f(y)x = \mathrm{f}(y) around the yy-axis over the interval from y=ay = a to y=by = b is ∫ab2πf(y)1+(f′(y))2 dy.\int_a^b 2\pi\mathrm{f}(y)\sqrt{1 + \left(\mathrm{f}'(y)\right)^2}\,\mathrm{d} y. Find numerical answer of the surface area of the spherical part of the light bulb. [4]

    3. Each light bulb can effectively light up a circular area of 0.50.5 m2^2. The art student wants to estimate the number of light bulbs needed for the art installation in part (a). The art student uses the following formula to calculate the number of light bulbs to acquire. Number of light bulbs=Area of Region R0.5.\text{Number of light bulbs} = \frac{\text{Area of Region } R}{0.5}. Given that 11 unit on the art installation represents 100100 m, determine the number of light bulbs that the art student would acquire. Suggest why the number of light bulbs using this formula may not be accurate. [2]

Show full worked solution▾

(a) ∫012ex−2x2e2−x dx=2e[x22]01−2∫01x2e2−x dx=e−{2[x2e2−x−1]01−2∫01e2−x−1(2x) dx}=e+2[x2e2−x]01−4∫01e2−x(x) dx=e+2[x2e2−x]01−4([xe2−x−1]01−∫01e2−x−1(1) dx)=e+2[x2e2−x]01+4[xe2−x]01−4[e2−x−1]01=11e−4e2\begin{aligned} \int_0^1 2\mathrm{e} x-2x^2\mathrm{e}^{2-x}\,dx &=2\mathrm{e}\left[\frac{x^2}{2}\right]_0^1-2\int_0^1x^2\mathrm{e}^{2-x}\,dx\\ &=\mathrm{e}-\left\{2\left[x^2\frac{\mathrm{e}^{2-x}}{-1}\right]_0^1-2\int_0^1\frac{\mathrm{e}^{2-x}}{-1}(2x)\,dx\right\}\\ &=\mathrm{e}+2\left[x^2\mathrm{e}^{2-x}\right]_0^1-4\int_0^1\mathrm{e}^{2-x}(x)\,dx\\ &=\mathrm{e}+2\left[x^2\mathrm{e}^{2-x}\right]_0^1-4\left(\left[x\frac{\mathrm{e}^{2-x}}{-1}\right]_0^1-\int_0^1\frac{\mathrm{e}^{2-x}}{-1}(1)\,dx\right)\\ &=\mathrm{e}+2\left[x^2\mathrm{e}^{2-x}\right]_0^1+4\left[x\mathrm{e}^{2-x}\right]_0^1-4\left[\frac{\mathrm{e}^{2-x}}{-1}\right]_0^1\\ &=11\mathrm{e}-4\mathrm{e}^2 \end{aligned} (b)(i) x2+(y−6)2=36x2=36−(y−6)2\begin{aligned} x^2+(y-6)^2&=36\\ x^2&=36-(y-6)^2 \end{aligned} When x=25x=2\sqrt5, (25)2+(y−6)2=36(2\sqrt5)^2+(y-6)^2=36

Then y=2y=2 or 1010 (reject)

Method 1

Volume of inert gas needed =π∫21236−(y−6)2 dy=π[36y−(y−6)33]212=8003π cm3\begin{aligned} &=\pi\int_2^{12}36-(y-6)^2\,dy\\ &=\pi\left[36y-\frac{(y-6)^3}{3}\right]_2^{12}\\ &=\frac{800}{3}\pi\ \text{cm}^3 \end{aligned} Method 2

Total volume of sphere 43π(6)3=288π\dfrac43\pi(6)^3=288\pi

Source marking note: formula for sphere not given, unlikely they will apply this method

Source marking note: Can provide as alternative.

Volume of the sphere that is not filled by inert gas =π∫0236−(y−6)2 dy=π[36y−(y−6)33]02=643π\begin{aligned} &=\pi\int_0^2 36-(y-6)^2\,dy\\ &=\pi\left[36y-\frac{(y-6)^3}{3}\right]_0^2\\ &=\frac{64}{3}\pi \end{aligned} Volume of inert gas in the light bulb =288π−643π=8003π cm3=288\pi-\frac{64}{3}\pi=\frac{800}{3}\pi\ \text{cm}^3

(ii) x=±36−(y−6)2x=\pm\sqrt{36-(y-6)^2}

Surface area of spherical part of light bulb =∫2122π(36−(y−6)2)1+(−2(y−6)236−(y−6)2)2 dy=376.991118\begin{aligned} &=\int_2^{12}2\pi\left(\sqrt{36-(y-6)^2}\right) \sqrt{1+\left(\frac{-2(y-6)}{2\sqrt{36-(y-6)^2}}\right)^2}\,dy\\ &=376.991118 \end{aligned} (iii) Since 1 unit =100=100 m, 1 unit2=10000^2=10000 m2^2.

Region R =(11e−4e2)(104)=3448.757=(11\mathrm{e}-4\mathrm{e}^2)(10^4)=3448.757 m2^2

Number of light bulbs the art student would acquire =3448.7570.5=6897.514≈6898=\frac{3448.757}{0.5}=6897.514\approx6898 The irregular shape of region R would result the light bulbs lighting up overlapping zones.

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