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Definite integrals, areas and volumes

Practice section 7 of 9

Questions 37–42 in this collection

Practice questions

Review practice · 2 of 6

Derive ellipse area, ellipsoid volume and sphere surface area

(a) The ellipse x2/a2+y2/b2=1x^2/a^2+y^2/b^2=1 has b>a>0b>a>0.

(i) Show its enclosed area is A=4ba∫0aa2−x2 dxA=\tfrac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx. [1] Use x=asin⁡θx=a\sin\theta to find AA in terms of a,b,πa,b,\pi. [4]

(ii) Rotate the enclosed region through π\pi radians about the x-axis to form an ellipsoid. Find its volume. [3]

(b) The curved surface area generated by completely rotating a nonnegative curve y=f(x)y=f(x), a≤x≤ba\le x\le b, about the x-axis is

S=∫ab2πy1+(dy/dx)2 dx.S=\int_a^b2\pi y\sqrt{1+(dy/dx)^2}\,dx.

Use the circle x2+y2=r2x^2+y^2=r^2, −r≤x≤r-r\le x\le r, to show the surface area of a sphere is 4πr24\pi r^2. [4]

[2019/ACJC/II/3]

CJC · Review problems · Question 5

Solution

Review practice · 3 of 6

Compare an arcsine integral with an absolute-value area

(a) Find the exact value of ∫12x2ln⁡x dx\int_1^2x^2\ln x\,dx. [3]

(b) Find ∫9/3+2x−x2 dx\int9/\sqrt{3+2x-x^2}\,dx. Hence find the exact constant aa satisfying

∫0293+2x−x2 dx=∫−1/2aπ∣1−2x∣ dx.\int_0^2\frac9{\sqrt{3+2x-x^2}}\,dx=\int_{-1/2}^{a}\pi|1-2x|\,dx.

[7]

[2013/RI/I/12]

CJC · Review problems · Question 6

Solution

Review practice · 4 of 6

Reverse reciprocal-substitution bounds and reuse an integral

(a)(i) For a>1a>1, use u=1/xu=1/x to show ∫1adx/(1+x2)=∫1/a1du/(1+u2)\int_1^a dx/(1+x^2)=\int_{1/a}^1du/(1+u^2). [4]

(ii) Hence show tan⁡−1a+tan⁡−1(1/a)=k\tan^{-1}a+\tan^{-1}(1/a)=k, finding kk. [3]

(b)(i) Show ∫01x3/(1+x2) dx=1/2−(1/2)ln⁡2\int_0^1x^3/(1+x^2)\,dx=1/2-(1/2)\ln2. [2]

(ii) Without using a calculator, evaluate ∫01x2tan⁡−1x dx\int_0^1x^2\tan^{-1}x\,dx. [3]

[2013/JJC/I/12]

CJC · Review problems · Question 7

Solution

Review practice · 5 of 6

Combine inverse-sine integration with two rotation radii

(a) Find ∫sin⁡−1(3x) dx\int\sin^{-1}(3x)\,dx. [3]

(b) The curves C1,C2C_1,C_2 are y=1−(3/π)x2y=1-(\sqrt3/\pi)x^2 and y=13cos⁡xy=\tfrac13\cos x. Rotate the first-quadrant region bounded by them and the y-axis through 360∘360^\circ about the y-axis. Find the volume to 2 decimal places. [4]

[2018/HCI/Promo/3]

CJC · Review problems · Question 8

Solution

Review practice · 6 of 6

Use horizontal strips for an inverse-sine region

The diagram shows y=sin⁡−1(x/2)y=\sin^{-1}(x/2) for 0≤x≤20\le x\le2.

(i) Find the area of R1R_1, bounded by the curve, y=π/12y=\pi/12, y=π/6y=\pi/6 and the y-axis.

(ii) Find the volume when R1R_1 rotates through 2π2\pi about the x-axis.

(iii) Without using a calculator, find the exact area of R2R_2, bounded by the curve, x=1x=1, x=3x=\sqrt3 and the x-axis.

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Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.
Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.
Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.

[2015/DHS/I/9]

CJC · Review problems · Question 9

Solution

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