Class 12: Areas of Bounded Regions – Exercise 21.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Using integration, find the area of the following regions:}$

$\displaystyle \textbf{Question 1: }~\text{Find the area of the region in the first quadrant bounded by the parabola } \\ y=4x^{2}\text{ and the lines }x=0,\ y=1\text{ and }y=4.$
$\displaystyle \text{Answer:}$

$\displaystyle y=4x^2\text{ represents a parabola opening upwards, symmetric about the } \\ y\text{-axis with vertex at }(0,0)$
$\displaystyle y=1\text{ is a line parallel to the }x\text{-axis, cutting the parabola at }\left(-\tfrac12,1\right)\text{ and }\left(\tfrac12,1\right)$
$\displaystyle y=4\text{ is a line parallel to the }x\text{-axis, cutting the parabola at }(-1,1)\text{ and }(1,1)$
$\displaystyle x=0\text{ is the }y\text{-axis}$
$\displaystyle \text{Consider a horizontal strip of length }|x|\text{ and width }dy\text{ in the first quadrant}$
$\displaystyle \text{Area of the approximating rectangle }=|x|\,dy$
$\displaystyle \text{The strip moves from }y=1\text{ to }y=4$
$\displaystyle \text{The required area is the area enclosed by the curve in the first quadrant between }y=1\text{ and }y=4$
$\displaystyle \text{Area of the shaded region }=\int_{1}^{4} |x|\,dy$
$\displaystyle \text{Since }x>0\text{ in the first quadrant, }|x|=x$
$\displaystyle A=\int_{1}^{4} x\,dy$
$\displaystyle \text{From }y=4x^2,\ x=\sqrt{\tfrac{y}{4}}$
$\displaystyle A=\int_{1}^{4} \sqrt{\tfrac{y}{4}}\,dy$
$\displaystyle =\tfrac12\int_{1}^{4} \sqrt{y}\,dy$
$\displaystyle =\tfrac12\left[\tfrac{y^{3/2}}{3/2}\right]_{1}^{4}$
$\displaystyle =\tfrac12\times\tfrac23\left(4^{3/2}-1^{3/2}\right)$
$\displaystyle =\tfrac13(8-1)$
$\displaystyle =\tfrac73$
$\displaystyle \text{Hence, the area enclosed by the parabola in the first quadrant between }y=1\text{ and }y=4\text{ is }\tfrac73\text{ sq. units}$

$\displaystyle \textbf{Question 2: }~\text{Find the area of the region bounded by }x^{2}=16y,\ y=1,\ \\ y=4\text{ and the } y\text{-axis in the first quadrant.}$
$\displaystyle \text{Answer:}$

$\displaystyle x^2=16y\text{ represents a parabola with vertex at }(0,0)\text{ and symmetric about the }y\text{-axis}$
$\displaystyle y=1\text{ is a line parallel to the }x\text{-axis cutting the parabola at }(-4,1)\text{ and }(4,1)$
$\displaystyle y=4\text{ is a line parallel to the }x\text{-axis cutting the parabola at }(-8,4)\text{ and }(8,4)$
$\displaystyle \text{Consider a horizontal strip of length }|x|\text{ and width }dy$
$\displaystyle \text{Area of the approximating rectangle }=|x|\,dy$
$\displaystyle \text{The strip moves from }y=1\text{ to }y=4$
$\displaystyle \text{The required area is the area enclosed by the curve in the first quadrant between }y=1\text{ and }y=4$
$\displaystyle \text{Area of the shaded region }=\int_{1}^{4}|x|\,dy$
$\displaystyle \text{Since }x>0\text{ in the first quadrant, }|x|=x$
$\displaystyle A=\int_{1}^{4} x\,dy$
$\displaystyle \text{From }x^2=16y,\ x=\sqrt{16y}$
$\displaystyle A=\int_{1}^{4}\sqrt{16y}\,dy$
$\displaystyle =4\int_{1}^{4}\sqrt{y}\,dy$
$\displaystyle =4\left[\frac{y^{3/2}}{3/2}\right]_{1}^{4}$
$\displaystyle =\frac{8}{3}\left(4^{3/2}-1^{3/2}\right)$
$\displaystyle =\frac{8}{3}(8-1)$
$\displaystyle =\frac{56}{3}$
$\displaystyle \text{Hence, the area enclosed by the parabola in the first quadrant between }y=1\text{ and }y=4\text{ is }\frac{56}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 3: }~\text{Find the area of the region bounded by }x^{2}=4ay\text{ and its latus rectum.}$
$\displaystyle \text{Answer:}$

$\displaystyle x^2=4ay\text{ represents a parabola with vertex at }(0,0)\text{, opening upwards } \\ \text{and symmetric about the }y\text{-axis}$
$\displaystyle F(0,a)\text{ is the focus of the parabola and }y=a\text{ is its latus rectum}$
$\displaystyle \text{Consider a horizontal strip of length }|x|\text{ and width }dy\text{ in the first quadrant}$
$\displaystyle \text{Area of the approximating rectangle }=|x|\,dy$
$\displaystyle \text{The strip moves from }y=0\text{ to }y=a$
$\displaystyle \text{Area }OAB=\int_{0}^{a}|x|\,dy$
$\displaystyle \text{Area }OAA'O=2\times\text{Area }OAB$
$\displaystyle A=2\int_{0}^{a}|x|\,dy$
$\displaystyle \text{Since }x>0\text{ in the first quadrant, }|x|=x$
$\displaystyle A=2\int_{0}^{a}x\,dy$
$\displaystyle \text{From }x^2=4ay,\ x=\sqrt{4ay}$
$\displaystyle A=2\int_{0}^{a}\sqrt{4ay}\,dy$
$\displaystyle =4\sqrt{a}\int_{0}^{a}\sqrt{y}\,dy$
$\displaystyle =4\sqrt{a}\left[\frac{y^{3/2}}{3/2}\right]_{0}^{a}$
$\displaystyle =\frac{8\sqrt{a}}{3}\left(a^{3/2}-0\right)$
$\displaystyle =\frac{8}{3}a^2$
$\displaystyle \text{Hence, the area enclosed by the parabola and its latus rectum is }\frac{8}{3}a^2\text{ sq. units}$

$\displaystyle \textbf{Question 4: }~\text{Find the area of the region bounded by }x^{2}+16y=0\text{ and its latus rectum.}$
$\displaystyle \text{Answer:}$

$\displaystyle x^2+16y=0\Rightarrow x^2=-16y$
$\displaystyle \text{Comparing with the standard parabola }x^2=4ay,\ a=-4$
$\displaystyle \text{Thus, }x^2+16y=0\text{ represents a parabola opening downwards with vertex at }O(0,0)$
$\displaystyle \text{The negative }y\text{-axis is the axis of symmetry}$
$\displaystyle \text{The focus of the parabola is }F(0,-4)$
$\displaystyle y=-4\text{ is the latus rectum of the parabola}$
$\displaystyle \text{The latus rectum cuts the parabola at }B(8,-4)\text{ and }B'(-8,-4)$
$\displaystyle x=8\text{ cuts the }x\text{-axis at }A(8,0)$
$\displaystyle \text{Area of the curve bounded by the latus rectum }=\text{shaded area }BOB'B=2(\text{Area }OBF)\ldots(1)$
$\displaystyle \text{Consider a vertical strip of length }|y|\text{ and width }dx\text{ in shaded area }OAB$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The strip moves from }x=0\text{ to }x=8$
$\displaystyle \text{Area of the shaded region }OAB=\int_{0}^{8}|y|\,dx$
$\displaystyle \text{From }x^2=-16y,\ y=-\frac{x^2}{16}$
$\displaystyle |y|=\frac{x^2}{16}$
$\displaystyle \text{Area of the shaded region }OAB=\int_{0}^{8}\frac{x^2}{16}\,dx$
$\displaystyle =\frac{1}{16}\left[\frac{x^3}{3}\right]_{0}^{8}$
$\displaystyle =\frac{8^3}{48}=\frac{32}{3}\text{ sq. units}\ldots(2)$
$\displaystyle \text{Area of rectangle }OABF=OA\times AB=8\times4=32\text{ sq. units}\ldots(3)$
$\displaystyle \text{From (2) and (3), Area }OBF=32-\frac{32}{3}=\frac{64}{3}\text{ sq. units}$
$\displaystyle \text{Shaded area }BOB'B=2\times\frac{64}{3}=\frac{128}{3}\text{ sq. units}$
$\displaystyle \text{Hence, the area of the curve }x^2+16y=0\text{ bounded by its latus rectum is }\frac{128}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 5: }~\text{Find the area of the region bounded by the curve }ay^{2}=x^{3},\ \text{the } \\ y\text{-axis and the lines }y=a\text{ and }y=2a.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation of the given curve is }ay^2=x^3$
$\displaystyle \text{The curve passes through the origin and is symmetrical about the }x\text{-axis}$
$\displaystyle \text{The lines }y=a\text{ and }y=2a\text{ are parallel to the }x\text{-axis}$
$\displaystyle \text{They intersect the }y\text{-axis at }(0,a)\text{ and }(0,2a)\text{ respectively}$
$\displaystyle \text{The required area is the area of the shaded region}$
$\displaystyle \text{Consider a horizontal strip of width }dy$
$\displaystyle \text{From }ay^2=x^3,\ x=(ay^2)^{1/3}$
$\displaystyle \text{Area of the strip }=x\,dy$
$\displaystyle \text{Required area }=\int_{a}^{2a} x\,dy$
$\displaystyle =\int_{a}^{2a}(ay^2)^{1/3}\,dy$
$\displaystyle =a^{1/3}\int_{a}^{2a}y^{2/3}\,dy$
$\displaystyle =a^{1/3}\left[\frac{y^{5/3}}{5/3}\right]_{a}^{2a}$
$\displaystyle =\frac{3}{5}a^{1/3}\left((2a)^{5/3}-a^{5/3}\right)$
$\displaystyle =\frac{3}{5}\left(2^{5/3}-1\right)a^2$
$\displaystyle \text{Hence, the required area }=\frac{3}{5}\left(2^{5/3}-1\right)a^2\text{ square units}$