Question
The diagram shows the curve with equation\( y =\sqrt{(1+3cos^{2}(\frac{1}{2}))}\)for 0 ≤ x ≤ 0. The region R is bounded by the curve, the axes and the line x =\( \pi\) .
(i) Use the trapezium rule with two intervals to find an approximation to the area of R, giving your answer correct to 3 significant figures.
(ii) The region R is rotated completely about the x-axis. Without using a calculator, find the exact volume of the solid produced.
Answer/Explanation
6(i) Use y values 2 \(\sqrt{2.5}\) , 1 or equivalents
Use correct formula, or equivalent, with\( h=\frac{1}{2}\pi \) and three y values
Obtain\( \frac{1}{2}\times \frac{1}{2}\pi (2+\sqrt[2]{2.5}+1)\) or equivalent and hence 4.84
6(ii) State or imply volume is \(\int \pi (1+3cos^{2}\frac{1}{2}x)dx\)
Use appropriate identity to express integrand in form \(k_{1}+k_{2}cosx\)
Obtain \(\int \pi (\frac{5}{2}+\frac{3}{2}cosx)dx or \int (\frac{5}{2}x+\frac{3}{2}cosx)dx\)
Integrate to obtain\( \pi (\frac{5}{2}x+\frac{3}{2}sinx) or \frac{5}{2}x+\frac{3}{2}x\)
Obtain\( \frac{5}{2}\pi ^{2}\) with no errors seen
Question
The diagram shows the line y=1 and the part of the curve \(y=\frac{2}{\sqrt{\left ( x+1 \right )}}\).
(i)Show that the equation \(y=\frac{2}{\sqrt{\left ( x+1 \right )}}\) can be written in the form \(x=\frac{4}{y^{2}}-1\)
(ii)Find \(\int \left ( \frac{4}{y^{2}}-1 \right )dy\).Hence find the area of the shaded region.
(ii)The shaded region is rotated through about the y-axis.Find the exact value of the volume of revolution obtained.
Answer/Explanation
(i) \(x=\frac{4}{y^{2}}-1\)
(ii)\(\int \left ( \frac{4}{y^{2}}-1 \right )dy=\left [- \frac{4}{y} -y\right ]\)
\(^{2}_{1}\left [ -\frac{4}{2}-2-\left ( -4-1 \right ) \right ]\)
(iii)\(\pi \int x^{2}dy=\left ( \pi \right )\int \left ( \frac{16}{y^{4}}-\frac{8}{y^{2}} +y\right )\)
\(\left ( \pi \right )\left [ -\frac{16}{3y^{3}}+\frac{8}{y} +y\right ]\)
\(\left ( \pi \right )\left [\left ( -\frac{16}{24}+4+2 \right )-\left ( -\frac{-16}{3}+8+1\right )\right ]\)
\(\frac{5\pi }{3}\)
Question
The diagram shows parts of the curves \(y=\left ( 2x-1 \right )^{2}\) and \(y^{2}=1-2x\), intersecting at points A AND b.
(i) State the coordinates of A
(ii)Find,showing all necessary working, the area of the shaded region.
Answer/Explanation
(i)\(A=\left ( \frac{1}{2},0 \right )\)
(ii)\( \int \left ( 1-2x \right )^{\frac{1}{2}}dx=\left [ \frac{\left ( 1-2x \right )^{\frac{3}{2}}}{\frac{3}{2}} \right ]\left [ \div \left ( -2 \right )\right ]\)
\(\int \left ( 2x-1 \right ) ^{2}dx=\left [ \frac{\left ( 2x-1 \right )^{3}}{3} \right ]\left [ \div 2 \right ]\)
\(\left [ 0-\left ( -\frac{1}{3} \right ) \right ]-\left [ 0-\left (- \frac{1}{6} \right ) \right ]\)
\(\frac{1}{6}\)
Question
The diagram shows part of the curve \(y=\frac{1}{2}\left ( x^{4}-1 \right )\),defined for\( x\geq 0\)
(i)Find, showing all necessary working ,the area of shaded region.
(ii)Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360^{\circ}\) about the axis.
(iii)Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360^{\circ}\) about the y-axis.
Answer/Explanation
(i)Area=\(\int \frac{1}{2}\left ( x^{4} -1\right )dx=\frac{1}{2}\left [ \frac{x^{5}}{5} -x\right ]\)
\(\frac{1}{2}\left [ \frac{1}{5}-1 \right ]-0=-\left (\frac{2}{5} \right )\)
(ii)\(Vol=\pi \int y^{2}dx=\frac{1}{4}\int \left ( x^{8}-2x^{4}+1 \right )dx\)
\(\frac{1}{4}\left ( \pi \right )\left [ \frac{x^{9}}{9}-\frac{2x^{5}}{5}+x \right ]\)
\(\frac{1}{4}\left ( \pi \right )\left [ \frac{1}{9}-\frac{2}{5}+1 \right ]-0\)
\(\frac{8\pi }{45}\) or 0.559
(iii)\(Vol=\pi \int x^{2}dy=\left ( \pi \right )\int \left ( 2y+1` \right )^{\frac{1}{2}}dy\)
\(\pi \left [ \frac{\left ( 2y+1 \right )^{\frac{3}{2}}}{\frac{3}{2}} \right ]\left [ \div 2 \right ]\)
\(\left ( \pi \right )\left [ \frac{1}{3} -0\right ]\)
\(\frac{\pi }{3}\) or 1.05
Question
The diagram shows part of the curve \(\left ( 1+4x \right )^{\frac{1}{2}} \) and a point P (6, 5) lying on the curve. The line PQ intersects the x-axis at Q (8, 0).
(i) Show that PQ is a normal to the curve.
(ii) Find, showing all necessary working, the exact volume of revolution obtained when the shaded region is rotated through \(360^{\circ}\) about the x-axis. In part (ii) you may find it useful to apply the fact that the volume, V, of a cone of base radius r and vertical height h, is given by \(V=\frac{1}{3}\pi r^{2}h\).
Answer/Explanation
(i)\(\frac{\mathrm{d} y}{\mathrm{d} x}=\left [ \frac{1}{2}(1+4x)^{-\frac{1}{2}} \right ]\times 4\)
At x=6 ,\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{5}
Gradient of PQ\(=-\frac{5}{2}\) hence PQ is a normal or \(m_{1}m_{2}=-1\)
(ii)Volume for curve=\(\pi \int (1+4x)\) and attempt to integrate \(y^{2}\)
\(=\pi \left [ x+2x^{2} \right ]\)
\(=\pi \left [ 6+72-0 \right ]\)
\(=78(\pi )\)
Vol for line \(=\frac{1}{3}\times \pi \times 5^{2}\times 2\)
\(=\frac{50}{3}\pi \)
Total volume \(=78\pi +\frac{50\pi }{3}=94\tfrac{2}{3}\pi \)
Question
The diagram shows the part of the curve \(y=\frac{4}{5-3x}\)
(i)Find the equation of the normal to the curve at the curve at the point where x=1 in th e form y=mx+c,where m and c are constants.
The shaded region is bounded by the curve,the coordinate axes and the line x=1
(ii)Find, showing all the necessary w2orking,the volume obtained when this shaded region is rotated through \(360^{\circ}\) about the axis
Answer/Explanation
.(i) \(\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{4}{\left ( 5-3x \right )^{2}}\times \left ( -3 \right )\)
Gradient of tangent=3,Gradient of normal\(=\frac{1}{3}\)
→eqn:\(y-2=-\frac{1}{3}\left ( x-1 \right )\)
→\(y=-\frac{1}{3}x+\frac{7}{3}\)
(ii)Vol\(=\pi \int_{0}^{1}\frac{16}{\left ( 5-3x \right )^{2}}dx\)
\(\pi \left [ \frac{-16}{\left ( 5-3x \right )}\div -3 \right ]\)
\(\left ( \pi \left ( \frac{16}{6}-\frac{16}{15} \right ) \right )=\frac{8\pi }{5}\) (if limits switched must show – to +)
Question
Solve the equation \(3 sin^2\Theta = 4 cos \Theta -1\) for 0° < 0 ≤ 360°.
Answer/Explanation
\(3\sin ^{2}\Theta =4\cos \Theta -1\)
Uses \(s^{2}+c^{2}=1\)
\(\rightarrow 3c^{2}+4c-4\left ( =0 \right )\)
\(\left ( \rightarrow c=\frac{2}{3} or -2\right )\)
\(\rightarrow \Theta =48.2^{\circ}\) or \(311.8^{\circ}\)
0.841,5.44 rads,A1 only
\(\left ( 0.268\pi ,1.73\pi \right )\)
Question
The diagram shows part of the curve with equation \(y=x^2+1\). The shaded region enclosed by the curve, the y-axis and the line y = 5 is rotated through \(360^0\) about the y-axis
Find the volume obtained.
Answer/Explanation
Ans:
\((\pi)\int (y-1)dy\)
\((\pi)[\frac{y^2}{2}-y]\)
\((\pi)[(\frac{25}{2}-5)-(\frac{1}{2}-1)]\)
\(8\pi\) or AWRT 25.1
Question.
The diagram shows part of the curve \(y =\frac{8}{x+2}\) and the line \(2y + x = 8\), intersecting at points A and B. The point C lies on the curve and the tangent to the curve at C is parallel to AB.
(a) Find, by calculation, the coordinates of A, B and C.
(b) Find the volume generated when the shaded region, bounded by the curve and the line, is rotated through \(360^{\circ}\) about the x-axis.
Answer/Explanation
(a) Simultaneous equations \(\frac{8}{x+2}=4=\frac{1}{2}x\)
x = 0 or x = 6→ A (0, 4) and B (6, 1)
At \(C=\frac{-8}{(x+2)^{2}}=-\frac{1}{2}\rightarrow C(2,2)\)
(B1 for the differentiation. M1 for equating and solving)
(b) Volume under line \(\pi \int (-\frac{1}{2}x+4)^{2}dx=\left [ \frac{x^{3}}{12}-2x^{2}+16x \right ]=(42\pi )\)
(M1 for volume formula. A2,1 for integration)
Volume under curve \(=\pi \int \left ( \frac{8}{x+2} \right )^{2}dx=\pi \left [ \frac{-64}{x+2} \right ]=(24\pi )\)
Subtracts and uses 0 to 6→ 18π