Q.Find the area of the region bounded by the line $x = 2$ and the parabola $y^2 = 8x$.

Showing the 12 most recent of 14 on this concept.

1. GUJCET 2022Set 081 markMCQ
   Q.The area of the region bounded by the curve $y^2=4x$ and the line $x=3$ is ______. (A) $3\sqrt{3}$ (B) $3\sqrt{8}$ (C) $8$ (D) $8\sqrt{3}$

   ›Reveal solutionSolution

   Integrate the full width $2\sqrt{4x}$ from $0$ to $3$.

   Concept. For $y^2=4x$, $y=\pm2\sqrt x$, so width $=4\sqrt x$.

   $A=\int_0^3 4\sqrt x\,dx=4\cdot\frac{2}{3}x^{3/2}\Big|_0^3=\frac{8}{3}\cdot3^{3/2}=\frac{8}{3}\cdot3\sqrt3=8\sqrt3.$

   ✓Final answer

   (D) $8\sqrt{3}$

   ANSWER: (D)

2. GUJCET 2020Set 071 markMCQ
   Q.The area of the parabola $y^2 = 4ax$ bounded by its latus rectum is ________. (A) $\frac{16}{3}a^2$ (B) $\frac{4}{3}a^2$ (C) $\frac{8}{3}a^2$ (D) $4a^2$

   ›Reveal solutionSolution

   Integrate $y=2\sqrt{ax}$ from $x=0$ to $x=a$ and double it (symmetry about the axis).

   Concept: The latus rectum is the line $x=a$. Area between the parabola and this chord:

   $A=2\int_0^a \sqrt{4ax}\,dx=2\int_0^a 2\sqrt a\,\sqrt x\,dx=4\sqrt a\left[\frac{2}{3}x^{3/2}\right]_0^a$

   $=4\sqrt a\cdot\frac{2}{3}a^{3/2}=\frac{8}{3}a^2$

   ✓Final answer

   (C) $\dfrac{8}{3}a^2$

   ANSWER: (C)

3. GUJCET 2023Set 091 markMCQ
   Q.Area of the region enclosed by the parabola $y = x^{2}$ and the line $y = x + 2$ is : (A) $\frac{9}{2}$ (B) $\frac{11}{2}$ (C) $\frac{5}{2}$ (D) $\frac{7}{2}$

   ›Reveal solutionSolution

   Integrate (line − parabola) between their intersection points.

   Concept. Intersections of $y=x^2$ and $y=x+2$: $x^2-x-2=0\Rightarrow x=-1,2$.

   Solution.

   $\int_{-1}^{2}\big[(x+2)-x^2\big]\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20+7}{6}=\frac{27}{6}=\frac{9}{2}.$

   ✓Final answer

   (A) $\frac{9}{2}$

   ANSWER: (A)

4. GUJCET 2022Set 081 markMCQ
   Q.The area of the region bounded by the two parabolas $y=x^2$ and $y^2=x$ is ______. (A) $\frac{3}{4}$ (B) $3$ (C) $\frac{1}{2}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   The curves meet at $(0,0)$ and $(1,1)$; integrate $\sqrt x-x^2$.

   Concept. Between $x=0$ and $x=1$, $y^2=x$ (upper, $y=\sqrt x$) lies above $y=x^2$.

   $A=\int_0^1\left(\sqrt x-x^2\right)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_0^1=\frac23-\frac13=\frac13.$

   ✓Final answer

   (D) $\frac{1}{3}$

   ANSWER: (D)

5. GUJCET 2022Set 081 markMCQ
   Q.The area of the parabola $x^2=12y$ bounded by its latus rectum is ______. (A) $3$ (B) $\frac{24}{3}$ (C) $24$ (D) $\frac{8}{3}$

   ›Reveal solutionSolution

   Here $4a=12\Rightarrow a=3$; the latus rectum is $y=3$, and the enclosed area is $24$.

   Concept. $x^2=12y$ has $a=3$; latus rectum at $y=3$, where $x=\pm6$. The region between the parabola and the line:

   $A=\int_{-6}^{6}\left(3-\frac{x^2}{12}\right)dx=2\left[3x-\frac{x^3}{36}\right]_0^6=2(18-6)=24.$

   ✓Final answer

   (C) $24$

   ANSWER: (C)

6. GUJCET 2021Set 151 markMCQ
   Q.Area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$, $x = 4$ and X-axis in the first quadrant is $\underline{\hspace{2cm}}$ (A) $\frac{14}{3}$ (B) $\frac{7}{3}$ (C) $\frac{28}{3}$ (D) $14$

   ›Reveal solutionSolution

   Integrate $y=\sqrt{x}$ from $x=1$ to $x=4$.

   Concept — area under a curve. In the first quadrant $y^2=x$ gives $y=\sqrt{x}$. The region is bounded by $x=1$, $x=4$ and the X-axis:

   $A=\int_1^4\sqrt{x}\,dx=\frac{2}{3}\Big[x^{3/2}\Big]_1^4=\frac{2}{3}(8-1)=\frac{14}{3}.$

   ✓Final answer

   (A) $\frac{14}{3}$

   ANSWER: (A)

7. GUJCET 2021Set 151 markMCQ
   Q.Area of the region bounded by the curve $y^2 = 4x$, Y-axis and the line $y = 3$ is $\underline{\hspace{2cm}}$ (A) $2$ (B) $\frac{9}{3}$ (C) $\frac{9}{4}$ (D) $\frac{9}{2}$

   ›Reveal solutionSolution

   Integrate along the y-axis: $x=y^2/4$ from $y=0$ to $y=3$.

   Concept — area bounded by a parabola and the Y-axis. From $y^2=4x$, $x=\frac{y^2}{4}$. The region is bounded by the Y-axis and the line $y=3$:

   $A=\int_0^3 x\,dy=\int_0^3\frac{y^2}{4}\,dy=\frac14\cdot\frac{y^3}{3}\Big|_0^3=\frac{27}{12}=\frac{9}{4}.$

   ✓Final answer

   (C) $\frac{9}{4}$

   ANSWER: (C)

8. GUJCET 2024Set 131 markMCQ
   Q.Area of the region bounded by the curve $x^2 = 4y$, X-axis and the line $x = 3$ is __________. (A) $\frac{9}{2}$ (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $2$

   ›Reveal solutionSolution

   Area $=\int_0^3 y\,dx$ with $y=x^2/4$.

   Concept. From $x^2=4y$, $y=\dfrac{x^2}{4}$. Area between the curve, the X-axis and $x=3$:

   $A=\int_0^3 \dfrac{x^2}{4}\,dx=\dfrac14\left[\dfrac{x^3}{3}\right]_0^3=\dfrac14\cdot\dfrac{27}{3}=\dfrac{9}{4}.$

   ✓Final answer

   Option (B) $\dfrac{9}{4}$

   ANSWER: (B)

9. GUJCET 2025Set 031 markMCQ
   Q.Area of the region bounded by the curve $x^2 = 4y$ and the line $y = 3$ is (A) $4\sqrt{3}$ (B) $2\sqrt{3}$ (C) $\sqrt{3}$ (D) $3\sqrt{3}$

   ›Reveal solutionSolution

   [!TLDR] $R^2 + (2\sqrt2 R)^2 = 9R^2$, $(9R^2)^{3/2}=27R^3$ $\Rightarrow$ ratio $= 27:1$.

   Concept

   Magnetic field at the centre of a ring is $B_c = \dfrac{\mu_0 I}{2R}$ and on its axis at distance $x$ is $B_a = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ (NCERT Moving Charges and Magnetism).

   Solution

   With $x = 2\sqrt{2}R$:

   $R^2 + x^2 = R^2 + 8R^2 = 9R^2, \qquad (9R^2)^{3/2} = 27R^3.$

   $B_a = \frac{\mu_0 I R^2}{2\,(27R^3)} = \frac{\mu_0 I}{54R}.$

   $\frac{B_c}{B_a} = \frac{\mu_0 I/(2R)}{\mu_0 I/(54R)} = \frac{54}{2} = 27.$

   So the ratio is $27 : 1$.

   [!ANSWER] (A)

   Note

   This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.

10. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
    Q.The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is:

    (a) $128/3$

    (b) $256/3$

    (c) $64/3$

    (d) $512/3$

    ›Reveal solutionSolution

    The parabola meets the line $y=16$ at $x=\pm4$; integrate $(16-x^2)$ between these limits.

    $x^2=16\Rightarrow x=\pm4$. By symmetry,

    Area $=2\displaystyle\int_0^4(16-x^2)\,dx=2\left[16x-\frac{x^3}{3}\right]_0^4=2\left(64-\frac{64}{3}\right)=2\cdot\frac{128}{3}=\frac{256}{3}$.

    ✓Final answer

    The correct option is (b) $256/3$.

11. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
    Q.Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is ______.

    (a) $\frac{9}{2}$

    (b) $3$

    (c) $\frac{9}{4}$

    (d) $2$

    ›Reveal solutionSolution

    Integrate with respect to $y$ since the region is naturally bounded by the y-axis and a horizontal line.

    From $y^2=4x$, $x=\frac{y^2}{4}$. Area $=\int_0^3 x\,dy=\int_0^3\frac{y^2}{4}dy=\left[\frac{y^3}{12}\right]_0^3=\frac{27}{12}=\frac{9}{4}$.

    ✓Final answer

    The correct option is (c) $\frac{9}{4}$.

12. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.The area of the parabola $y^2 = 12x$ bounded by its latus rectum is ___.

    (a) $24$

    (b) $12$

    (c) $18$

    (d) $30$

    ›Reveal solutionSolution

    Integrate the parabola up to its latus rectum $x=a=3$ and double for symmetry.

    $y^2 = 12x \Rightarrow 4a=12 \Rightarrow a=3$; latus rectum is at $x=3$.

    Area $= 2\int_0^3 \sqrt{12x}\,dx = 2\cdot 2\sqrt3\int_0^3 x^{1/2}\,dx = 4\sqrt3\left[\tfrac{2}{3}x^{3/2}\right]_0^3.$

    $= 4\sqrt3 \cdot \tfrac23 \cdot 3\sqrt3 = 4\sqrt3\cdot 2\sqrt3 = 8\cdot 3 = 24.$

    ✓Final answer

    (a) $24$.