Application of Integrals Class 12 NCERT Solutions Chapter 8 Ex 8.1

Picture-first. Read the equation at sight: the semi-axes are a=√16=4 along the x-direction and b=√9=3 along the y-direction. The ellipse sits symmetrically about both axes, so we only need to integrate over one quadrant and multiply by 4.

Concept used. An ellipse in standard form has two mutually perpendicular axes of symmetry (the coordinate axes themselves), so the four quadrants contribute equally to the total area. The first-quadrant area is A_Q1=₀^ay dx, and the total area is A=4A_Q1.

1. Solve for y in the upper half. From the ellipse equation y²=9(1-x²16)=916(16-x²) y=34√16-x².
2. Set up the symmetric integral. The total area is A=4₀⁴34√16-x² dx =3₀⁴√16-x² dx.
3. Trigonometric substitution. Put x=4sinθ, so dx=4cosθ dθ and √16-x²=4cosθ. Limits change: x=0⇒θ=0; x=4⇒θ=π2. ₀⁴√16-x² dx =₀^π/2(4cosθ)(4cosθ) dθ =16₀^π/2cos²θ dθ.
4. Half-angle identity. Use cos²θ=1+cos 2θ2: 16₀^π/21+cos 2θ2 dθ =8[θ+sin 2θ2]₀^π/2. At θ=π2: π2+sinπ2=π2. At θ=0: 0+0=0. So the integral equals 8·π2=4π.
5. Multiply by 3. A=3× 4π=12π.

Why this matters. The same workflow - solve for the non-strip variable, exploit symmetry, evaluate the standard integral ∫√a²-x² dx - handles every circle / ellipse area question in this chapter. Memorise the structure, not just the answer.

A=12π square units.

Picture-first. A look at x²4+y²9=1 tells me the curve passes through (± 2,0) and (0,± 3). The ellipse is taller than wide, so I might equally have used horizontal strips (x in terms of y) - but the formula and the symmetry argument work identically.

Concept used. For a standard-form ellipse, the four quadrant pieces are congruent under reflection in the axes; the total area is therefore four times the first-quadrant area. The first-quadrant area can be found with vertical or horizontal strips.

1. Horizontal-strip set-up. Solving for x in the first quadrant: from x²4+y²9=1, x²=4(1-y²9)=4(9-y²)9 x=23√9-y².
2. Symmetric area integral. Total area A=4₀³x dy=4₀³23√9-y² dy =83₀³√9-y² dy.
3. Substitution. Put y=3sinφ, dy=3cosφ dφ; √9-y²=3cosφ. Limits: y=0φ=0, y=3φ=π2. ₀³√9-y² dy =₀^π/2(3cosφ)(3cosφ) dφ =9₀^π/2cos²φ dφ.
4. Half-angle. cos²φ=1+cos 2φ2: 9·12[φ+sin 2φ2]₀^π/2 =92[π2+0-0-0]=9π4.
5. Multiply by 83. A=83·9π4=72π12=6π.

Why this matters. Vertical strips and horizontal strips must give the same area, by construction. Switching between them is a powerful sanity check and often makes the algebra cleaner when one axis offers an easier substitution.

A=6π square units.

Quick reading. The line x=2 touches the circle of radius 2 at exactly one point, (2,0). The line x=0 is the y-axis, which meets the circle at (0,± 2). So the region trapped between the curves x=0, x=2, y=0 (which is implied by ``first quadrant'') and the arc of x²+y²=4 is the entire quarter-disk in the first quadrant.

Concept used. The disk (x,y):x²+y²≤ r² is symmetric under reflection in both axes, so its four quadrants have equal area. Total disk area is π r²; each quadrant therefore has area 14π r².

1. Identify the region as a quarter disk. First-quadrant arc of x²+y²=4 together with the x-axis segment [0,2] and the y-axis segment [0,2] traces the boundary of one quadrant of the disk.
2. Use the area formula. Quarter-disk area A=14 π r²=14 π (2)²=4π4=π.
3. Cross-check by integration. A=₀²√4-x² dx =[x2√4-x²+2sin^-1x2]₀² =π. Both methods agree.

Why this matters. In MCQs, recognising a region as a ``quarter-disk'', ``half-ellipse'', or ``30^∘ circular sector'' collapses pages of integration into a single multiplication. Always sketch first.

Option (A): A=π.

Strategic angle. The bounding lines x=0 and y=3 are parallel to the coordinate axes, and the parabola opens to the right. Picture a thin horizontal slab of thickness dy at height y: its left end sits on the y-axis (x=0), its right end on the parabola (x=y²/4). Length of slab =y²4-0=y²4.

Concept used. For a region whose left boundary is the y-axis and whose right boundary is a curve x=g(y), between horizontal lines y=c and y=d, A=_c^dg(y) dy. This is the horizontal-strip version of the fundamental area integral.

1. Boundaries. Bottom y=0 (the parabola meets the y-axis at the origin), top y=3 (given), left x=0, right x=y²4.
2. Set up. A=₀³(y²4-0)dy =14₀³y² dy.
3. Antiderivative. ∫ y² dy=y³3. Hence A=14[y³3]₀³ =14(273-0)=14(9)=94.
4. Verify by a sketch-check. The region is a thin ``sliver'' that widens with y; its area is positive and less than the bounding rectangle 0≤ x≤ 9/4, 0≤ y≤ 3 whose area is 94× 3=274=6.75. Our value 2.25 is well inside that bound .

Why this matters. The choice between ∫ y dx and ∫ x dy is rarely a matter of taste: the bounding curves dictate which one keeps the integrand a single function. Pick the strip direction that gives a single-formula integrand over the whole interval.

Option (B): A=94.