### 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.

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$$\frac{1}{2}\times \frac{1}{2}\pi (2+\sqrt{2.5}+1)$$ 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.

(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.

(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.

(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 $$V=\frac{1}{3}\pi r^{2}h$$.

(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

.(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°.

$$3\sin ^{2}\Theta =4\cos \Theta -1$$

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 $$311.8^{\circ}$$

$$\left ( 0.268\pi ,1.73\pi \right )$$

### 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.

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.

(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