Q.The area of the region bounded by the circle $x^2 + y^2 = 1$ is
(A) $2\pi$ sq units
(B) $\pi$ sq units
(C) $3\pi$ sq units
(D) $4\pi$ sq units

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The area of the region (in sq.units) bounded by the curves $x^2 + y^2 = 16$ and $y^2 = 6x$ is (A) $4\pi + 4\sqrt{3}$ (B) $\dfrac{2}{3}\left(4\pi + \sqrt{3}\right)$ (C) $\dfrac{4}{3}\left(4\pi + \sqrt{3}\right)$ (D) $\dfrac{4\pi+\sqrt{3}}{3}$

   ›Reveal solutionSolution

   The region common to the circle $x^2+y^2=16$ and parabola $y^2=6x$ is bounded by the parabola near the origin and the circle further out; integrating each piece and doubling for symmetry gives $\tfrac43(4\pi+\sqrt3)$.

   Concept and Intuition

   The parabola $y^2=6x$ opens rightward from the origin, and the circle has radius 4. Near the vertex, the parabola is the "narrower" curve (smaller $|y|$ for given $x$), so it bounds the common region; farther out, the circle becomes narrower and takes over as the boundary. The crossover is exactly at their intersection point.

   Step-by-Step Solution

   1. Intersection: substitute $y^2=6x$ into $x^2+y^2=16$: $x^2+6x-16=0\Rightarrow x=2$ or $x=-8$ (rejected, since $y^2=6x\ge0$ needs $x\ge0$). At $x=2$: $y^2=12\Rightarrow y=\pm2\sqrt3$.
   2. For $x\in[0,2]$: parabola gives smaller $|y|$ than the circle (check at $x=1$: parabola $|y|=\sqrt6\approx2.45$, circle $|y|=\sqrt{15}\approx3.87$) — so the parabola bounds the region here.
   3. For $x\in[2,4]$: circle gives smaller $|y|$ (check at $x=3$: parabola $|y|=\sqrt{18}\approx4.24$, circle $|y|=\sqrt7\approx2.65$) — circle bounds here.
   4. Area (using symmetry about the $x$-axis, factor 2):

   $A=2\left[\int_0^2\sqrt{6x}\,dx+\int_2^4\sqrt{16-x^2}\,dx\right].$

   5. $\displaystyle\int_0^2\sqrt{6x}\,dx=\sqrt6\cdot\dfrac23x^{3/2}\Big|_0^2=\sqrt6\cdot\dfrac23\cdot2\sqrt2=\dfrac{8\sqrt3}3$.
   6. Using $\int\sqrt{16-x^2}dx=\dfrac x2\sqrt{16-x^2}+8\sin^{-1}\dfrac x4$: …

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2. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The area of the region under the curve $y = |\sin x - \cos x|$, $0 \le x \le \dfrac{\pi}{2}$ and above x-axis, is (in square units) (A) $2\sqrt{2}$ (B) $2\sqrt{2} - 1$ (C) $2(\sqrt{2}-1)$ (D) $2(\sqrt{2}+1)$

   ›Reveal solutionSolution

   Split the region at $x=\pi/4$ where $\sin x=\cos x$, integrate each branch of the absolute value separately, and add.

   Concept and Intuition

   $|\sin x-\cos x|$ is $\cos x-\sin x$ for $x<\pi/4$ (where cosine dominates) and $\sin x-\cos x$ for $x>\pi/4$ (where sine dominates) — the area under an absolute-value curve must be computed piecewise across the sign change.

   Step-by-Step Solution

   1. $\sin x=\cos x$ at $x=\pi/4$ within $[0,\pi/2]$; for $x<\pi/4$, $\cos x>\sin x$, so $|\sin x-\cos x|=\cos x-\sin x$; for $x>\pi/4$, it's $\sin x-\cos x$.
   2. $\displaystyle\int_0^{\pi/4}(\cos x-\sin x)dx=\Big[\sin x+\cos x\Big]_0^{\pi/4}=\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1$. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The area (in sq. units) of the region bounded by the lines $x=0$, $x=\dfrac{\pi}{2}$ and $f(x)=\sin x$, $g(x)=\cos x$ is (A) $2(\sqrt2-1)$ (B) $2(\sqrt3-1)$ (C) $2(\sqrt2+1)$ (D) $3\sqrt2+1$

   ›Reveal solutionSolution

   The two curves $\sin x$ and $\cos x$ cross at $x=\pi/4$ inside $[0,\pi/2]$, so the enclosed area is the sum of two pieces, each evaluating to $\sqrt2-1$, giving total $2(\sqrt2-1)$.

   Concept and Intuition

   Since $\sin x$ and $\cos x$ swap which one is larger at $x=\pi/4$, the area between them over $[0,\pi/2]$ must be split at that crossing point and the absolute difference integrated on each side.

   Step-by-Step Solution

   1. On $[0,\pi/4]$: $\cos x\ge\sin x$, so area contribution is $\int_0^{\pi/4}(\cos x-\sin x)dx=[\sin x+\cos x]_0^{\pi/4}=\left(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1$. …

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4. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.The area bounded by $y-1=-|x|$ and $y+1=|x|$ is (A) $\frac{1}{2}$ (B) $1$ (C) $2$ (D) $0$

   ›Reveal solutionSolution

   The two absolute-value "V" graphs cross at $(\pm1,0)$ and enclose a rhombus-shaped region (a square rotated 45°) of area $2$.

   Concept and Intuition

   $y-1=-|x|\Rightarrow y=1-|x|$ is an upside-down V peaking at $(0,1)$ with slopes $\mp1$. $y+1=|x|\Rightarrow y=|x|-1$ is a right-side-up V bottoming at $(0,-1)$ with slopes $\pm1$. Since both have unit slopes, the enclosed figure is actually a square with diagonals along the axes (vertices at $(0,1),(1,0),(0,-1),(-1,0)$), i.e. a rhombus/square of diagonal length 2 each way.

   Step-by-Step Solution

   1. Find intersections: $1-|x|=|x|-1\Rightarrow 2=2|x|\Rightarrow|x|=1\Rightarrow x=\pm1$, giving points $(1,0)$ and $(-1,0)$.
   2. On $(-1,1)$, the top curve is $y=1-|x|$ (value $1$ at $x=0$) and the bottom curve is $y=|x|-1$ (value $-1$ at $x=0$).
   3. Area $=\int_{-1}^{1}\big[(1-|x|)-(|x|-1)\big]dx=\int_{-1}^1(2-2|x|)dx$. …

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5. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.The area bounded by the curves $y-1=\cos x$, $y=\sin x$ and the X-axis between $x=0$ and $x=\pi$ is (A) $2+\dfrac{\pi}{2}$ (B) $-\dfrac{\pi}{2}$ (C) $2-\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   The area of the region bounded by $y=1+\cos x$, $y=\sin x$ and the X-axis on $[0,\pi]$ is $\dfrac{\pi}{2}$.

   Concept and Intuition

   Three curves fence off one closed region above the X-axis. Its floor is $y=0$; its roof is whichever curve is lower at each $x$ (the lower envelope), because that is what actually caps the region touching the axis. So the area is the integral of $\min(1+\cos x,\ \sin x)$.

   Step-by-Step Solution

   1. Find the crossing: $1+\cos x=\sin x\Rightarrow \sin x-\cos x=1\Rightarrow \sqrt2\sin\!\left(x-\tfrac{\pi}{4}\right)=1$, giving $x=\tfrac{\pi}{2}$ and $x=\pi$.
   2. On $[0,\tfrac{\pi}{2}]$: at $x=0$, $\sin x=0<1+\cos x=2$, so $\sin x$ is the lower (roof) curve.
   3. On $[\tfrac{\pi}{2},\pi]$: at $x=\tfrac{3\pi}{4}$, $1+\cos x\approx0.29<\sin x\approx0.71$, so $1+\cos x$ is the roof.
   4. $\displaystyle\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}=1$.
   5. $\displaystyle\int_{\pi/2}^{\pi}(1+\cos x)\,dx=[x+\sin x]_{\pi/2}^{\pi}=\pi-\left(\tfrac{\pi}{2}+1\right)=\tfrac{\pi}{2}-1.$ …

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6. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The area (in sq. units) of the smaller region lying above the X-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is (A) $2a^2\left(\dfrac{\pi}{4}-\dfrac{2}{3}\right)$ (B) $a^2\left(\dfrac{\pi}{4}-\dfrac{2}{3}\right)$ (C) $a^2\left(\dfrac{\pi}{4}+\dfrac{2}{3}\right)$ (D) $a^2\left(\dfrac{\pi^2}{4}-\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   The circle and parabola meet at $x=0$ and $x=a$; integrating the gap between the (higher) circle and the (lower) parabola over $[0,a]$ gives the smaller enclosed area $a^2\left(\dfrac\pi4-\dfrac23\right)$.

   Concept and Intuition

   The circle $x^2+y^2=2ax$, i.e. $(x-a)^2+y^2=a^2$, is centered at $(a,0)$ with radius $a$, so it passes through the origin and through $(2a,0)$. The parabola $y^2=ax$ opens rightward from the origin. Near $x=0$ the circle bulges upward faster than the parabola (its upper-half slope near the origin is steeper), and they cross again at $x=a$ — the "smaller" lens-shaped region above the x-axis is exactly the strip between the two curves over $x\in[0,a]$.

   Step-by-Step Solution

   1. Find intersections: substitute $y^2=ax$ into $x^2+y^2=2ax$: $x^2+ax=2ax \Rightarrow x^2-ax=0 \Rightarrow x(x-a)=0 \Rightarrow x=0$ or $x=a$.
   2. At $x=a$: $y^2=a\cdot a=a^2 \Rightarrow y=\pm a$; take $y=a$ for the region above the x-axis. So the curves cross at $(0,0)$ and $(a,a)$.
   3. For $0<x<a$, compare the upper branches: circle gives $y=\sqrt{2ax-x^2}$, parabola gives $y=\sqrt{ax}$. Near $x=0^+$, $2ax-x^2\approx2ax > ax$, so the circle lies above the parabola throughout $(0,a)$ — this strip is the smaller enclosed region.
   4. Area $=\displaystyle\int_0^a\left[\sqrt{2ax-x^2}-\sqrt{ax}\right]dx$.
   5. For $\int_0^a\sqrt{ax}\,dx = \sqrt a\cdot\dfrac23x^{3/2}\Big|_0^a = \sqrt a\cdot\dfrac23a^{3/2}=\dfrac23a^2$. …

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7. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The area of the region bounded by the curve $xy = -a$ $(a > 1)$ and the lines $x = -a$ and $y = a$ is (A) $a(a - 1 - \log a)$ (B) $a(a + 1 + \log a)$ (C) $a^2 - a + \log a$ (D) $a^2 - a - \log a$

   ›Reveal solutionSolution

   The area enclosed by the rectangular hyperbola $xy=-a$ and the lines $x=-a$, $y=a$ works out, after a direct integration, to $a(a-1-\log a)$.

   Concept and Intuition

   $xy=-a$ ($a>0$) is a hyperbola lying in the second and fourth quadrants (since the product of coordinates must be negative). We only need the branch in the second quadrant here ($x<0,y>0$, i.e. $y=-a/x$). The two given lines pin down a finite region between the curve and the corner point where the lines would meet.

   Step-by-Step Solution

   1. Rewrite the curve as $y=-\dfrac{a}{x}$ (valid for $x<0$ here, giving $y>0$).
   2. Find where the curve meets $x=-a$: $y=-a/(-a)=1$, point $(-a,1)$.
   3. Find where the curve meets $y=a$: $a=-a/x\Rightarrow x=-1$, point $(-1,a)$.
   4. For $x\in[-a,-1]$, the curve $y=-a/x$ lies below the line $y=a$ (check at $x=-1$: curve value $=a$, equal; at $x=-a$: curve value$=1<a$ since $a>1$). So the vertical strip between the curve and the top line $y=a$, from $x=-a$ to $x=-1$, is exactly the bounded region.
   5. Area $=\displaystyle\int_{-a}^{-1}\Big[a-\big(-\tfrac{a}{x}\big)\Big]dx=\int_{-a}^{-1}\Big(a+\tfrac{a}{x}\Big)dx$. …

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8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The area of the region bounded by the curve $y = x^2 + x$, the lines $y = x$, $x = 1$ and $y = 2$ is (A) $\dfrac{12}{5}$ (B) $\dfrac{7}{2}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The four boundary curves pin down a single closed loop from $x=0$ to $x=1$ between the parabola and the line $y=x$; its area is $\dfrac13$.

   Concept and Intuition

   When several curves are said to "bound a region," first locate every pairwise intersection — the closed loop's vertices are exactly these intersection points, and its area is found by integrating (upper curve minus lower curve) over the right interval.

   Step-by-Step Solution

   1. Intersection of $y=x^2+x$ and $y=x$: $x^2+x=x\Rightarrow x^2=0\Rightarrow x=0$. They only touch at $(0,0)$, and since $x^2+x-x=x^2\ge0$, the parabola is above the line for all other $x$.
   2. Intersection of $y=x$ and $x=1$: point $(1,1)$.
   3. Intersection of $y=x^2+x$ and $x=1$: point $(1,2)$.
   4. Intersection of $y=x^2+x$ and $y=2$: $x^2+x-2=0\Rightarrow(x-1)(x+2)=0\Rightarrow x=1$ (the relevant root, giving $(1,2)$ again). …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The area (in sq. units) of the region bounded by the curves $y=4|\cos x|$ and $y=-|\cos x|$ from $x=-\pi/2$ to $\pi/2$ is (A) 6 (B) 8 (C) 12 (D) 10

   ›Reveal solutionSolution

   Since $\cos x\ge0$ throughout $[-\pi/2,\pi/2]$, both curves are simple cosine multiples, and the enclosed area is $5\times\int\cos x\,dx=10$.

   Concept and Intuition

   The area between $y=f(x)$ (top curve) and $y=g(x)$ (bottom curve) over an interval is $\int[f(x)-g(x)]\,dx$. Here the vertical gap between the two curves is constant multiple of $|\cos x|$, so no case-splitting on sign is even needed within this interval.

   Step-by-Step Solution

   1. On $x\in[-\pi/2,\pi/2]$, $\cos x\ge0\Rightarrow|\cos x|=\cos x$.
   2. Top curve: $y=4\cos x$; bottom curve: $y=-\cos x$.
   3. Vertical gap: $4\cos x-(-\cos x)=5\cos x\ge0$ throughout, so this is directly the integrand. …

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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.The Area (in sq. units) of the region bounded by $x = 0, x = \dfrac{\pi}{2}$, X-axis, $y = \cos x$ and $y = \tan x$ is (A) $\dfrac{\sqrt{5} - 1}{2} + \dfrac{1}{2}\log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (B) $\dfrac{3 - \sqrt{5}}{2} + \log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (C) $\dfrac{\sqrt{5} - 1}{2} - \log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (D) $\dfrac{3 - \sqrt{5}}{2} + \dfrac{1}{2}\log\left(\dfrac{\sqrt{5} + 1}{2}\right)$

    ›Reveal solutionSolution

    Since $\tan x\to\infty$ at $\pi/2$, the finite region is bounded above by the lower of $\cos x$ and $\tan x$; splitting the integral at their intersection point gives $\dfrac{3-\sqrt5}2+\dfrac12\log\dfrac{\sqrt5+1}2$.

    Concept and Intuition

    $\cos x$ and $\tan x$ cross exactly once in $(0,\pi/2)$. Since $\tan x$ diverges as $x\to\pi/2^-$, a region bounded by the upper envelope of the two curves would have infinite area; the sensible, finite area bounded by the $x$-axis and both curves on $[0,\pi/2]$ is the area under whichever curve is lower at each $x$ — i.e. under $\tan x$ before the crossing and under $\cos x$ after it.

    Step-by-Step Solution

    1. Find the crossing point: $\cos x=\tan x=\dfrac{\sin x}{\cos x}\Rightarrow\cos^2x=\sin x\Rightarrow 1-\sin^2x=\sin x\Rightarrow\sin^2x+\sin x-1=0$.

    $\sin x_0=\frac{-1+\sqrt5}{2}=s\quad(\text{taking the root in }[0,1]).$

    Note also $\cos^2x_0=\sin x_0=s$ (directly from the defining equation), so $\cos x_0=\sqrt s$.

    2. Near $x=0$: $\cos0=1>\tan0=0$, so $\cos x$ is the upper curve, $\tan x$ the lower, for $x\in(0,x_0)$.

    Near $x=\pi/2$: $\tan x\to\infty>\cos x\to0$, so $\tan x$ is upper, $\cos x$ lower, for $x\in(x_0,\pi/2)$.

    3. The finite bounded area is therefore

    $A=\int_0^{x_0}\tan x\,dx+\int_{x_0}^{\pi/2}\cos x\,dx.$

    4. First piece: $\displaystyle\int_0^{x_0}\tan x\,dx=\Big[-\log(\cos x)\Big]_0^{x_0}=-\log(\cos x_0)=-\log\sqrt s=-\tfrac12\log s$.
    5. Second piece: $\displaystyle\int_{x_0}^{\pi/2}\cos x\,dx=\Big[\sin x\Big]_{x_0}^{\pi/2}=1-\sin x_0=1-s$. …

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11. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.If the area of the region enclosed by the curve $x^2 + y^2 = 16$ and the lines $x = 2$ and $x = 3$ is $\left(3\sqrt{7} - 4\sqrt{3} - \dfrac{8\pi}{3} + k\right)$ sq.units, then 'k' equals ______ (A) $16\sin^{-1}\left(\dfrac{3}{4}\right)$ (B) $8\sin^{-1}\left(\dfrac{3}{4}\right)$ (C) $4\sin^{-1}\left(\dfrac{3}{4}\right)$ (D) $2\sin^{-1}\left(\dfrac{3}{4}\right)$

    ›Reveal solutionSolution

    Compute the area between $x=2$ and $x=3$ under the full circle (upper + lower) using the standard $\int\sqrt{r^2-x^2}\,dx$ formula, then match the constant term to identify $k$.

    Concept and Intuition

    The region bounded by the circle between two vertical lines $x=2$ and $x=3$ (with both halves counted) is twice the area under the upper semicircle over that range — a direct application of the standard circle-area antiderivative.

    Step-by-Step Solution

    1. Area $=2\displaystyle\int_2^3\sqrt{16-x^2}\,dx$ (factor 2 for upper + lower half).
    2. Antiderivative: $\displaystyle\int\sqrt{16-x^2}\,dx=\dfrac{x}2\sqrt{16-x^2}+8\sin^{-1}\left(\dfrac x4\right)+c$.
    3. At $x=3$: $\dfrac32\sqrt7+8\sin^{-1}\left(\dfrac34\right)$.
    4. At $x=2$: $1\cdot\sqrt{12}+8\sin^{-1}\left(\dfrac12\right)=2\sqrt3+8\cdot\dfrac{\pi}6=2\sqrt3+\dfrac{4\pi}3$.
    5. Definite integral $=\dfrac32\sqrt7+8\sin^{-1}\left(\dfrac34\right)-2\sqrt3-\dfrac{4\pi}3$. …

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12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The area of the region lying between the curves $y = \sqrt{4-x^2}$, $y^2 = 3x$ and the Y-axis is (A) $\frac{\pi}{3} - \frac{1}{2\sqrt{3}}$ (B) $\frac{\pi}{6} + \frac{1}{2\sqrt{3}}$ (C) $\frac{\pi}{3} + \frac{1}{2\sqrt{3}}$ (D) $\frac{\pi}{6} - \frac{1}{2\sqrt{3}}$

    ›Reveal solutionSolution

    Integrating with respect to $y$ in two pieces — the parabola from $y=0$ to $\sqrt3$ and the circle from $y=\sqrt3$ to $2$ — gives the enclosed area as $\frac{\pi}{3}-\frac{1}{2\sqrt3}$.

    Concept and Intuition

    When a region's boundary is naturally expressed as $x=x(y)$ for both curves (here $x=\sqrt{4-y^2}$ for the circle and $x=y^2/3$ for the parabola), integrating along $y$ with the y-axis as the common left boundary is far cleaner than integrating along $x$.

    Step-by-Step Solution

    1. Find the intersection of $y=\sqrt{4-x^2}$ (so $y^2=4-x^2$) and $y^2=3x$: $3x=4-x^2\Rightarrow x^2+3x-4=0\Rightarrow(x+4)(x-1)=0$. Since $x\ge0$ on the parabola, $x=1$, giving $y=\sqrt3$.
    2. The circle meets the y-axis at $(0,2)$; the parabola meets it at the origin $(0,0)$.
    3. The enclosed region has: the y-axis as its left edge from $(0,0)$ to $(0,2)$; the parabola $x=y^2/3$ as its right edge for $y\in[0,\sqrt3]$; and the circle $x=\sqrt{4-y^2}$ as its right edge for $y\in[\sqrt3,2]$.
    4. Area $=\displaystyle\int_0^{\sqrt3}\frac{y^2}{3}\,dy+\int_{\sqrt3}^{2}\sqrt{4-y^2}\,dy$.
    5. First integral: $\displaystyle\int_0^{\sqrt3}\frac{y^2}{3}dy=\frac{1}{9}\left[y^3\right]_0^{\sqrt3}=\frac{3\sqrt3}{9}=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}$. …

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