JEE Main & Advanced

Application of Integrals

Integral Calculus: Applications of Integrals

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Module 1

Planar Quadrature & Area Enclosures

Area Under a Curve and Vertical vs. Horizontal Strip MechanicsTopic 1

Planar quadrature utilizes definite integration to evaluate the geometric area enclosed by continuous curves. For a continuous function $y = f(x)$ bounded between the vertical lines $x = a$ and $x = b$, the net physical area bounded between the curve and the x-axis requires the absolute value configuration to prevent signed cancellation: \[ A = \int_a^b |f(x)| \, dx \] The evaluation can be approached using two distinct geometric setups depending on the orientation of the differential slicing strips:
  • Vertical Slicing Strips (Integration with respect to $x$): Used when functions are naturally given as $y = f(x)$. The strip has width $dx$ and height $y = f(x)$. The area is $A = \int_a^b f(x) \, dx$.
  • Horizontal Slicing Strips (Integration with respect to $y$): Essential when functions are easier to express as $x = g(y)$ or when a curve contains a horizontal boundary trace across the y-axis bounded by $y = c$ and $y = d$. The strip has width $dy$ and horizontal length $x = g(y)$. The area is: \[ A = \int_c^d |g(y)| \, dy \]
A common trap in competitive exams is integrating blindly across areas where a function dips below the axis or changes its definition, which yields a wrong arithmetic result if absolute slicing is neglected.
Worked Examples
1

Find the area enclosed by the parabola $y^2 = 4x$ and the vertical line $x = 4$ using vertical strips, and verify the result using horizontal strips.

Show solution
The parabola is symmetric about the x-axis, extending from $x = 0$ to $x = 4$.
  • Method 1: Vertical Strip Slicing
    The upper half of the parabola is given by $y = 2\sqrt{x}$. Due to symmetry, the total area is twice the area of the upper region: \[ A = 2 \int_0^4 2\sqrt{x} \, dx = 4 \int_0^4 x^{1/2} \, dx \] Integrate using the standard power rule: \[ A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_0^4 = 4 \cdot \frac{2}{3} \cdot \left[ 4^{3/2} - 0 \right] = \frac{8}{3} \cdot 8 = \frac{64}{3} \]
  • Method 2: Horizontal Strip Slicing
    When $x = 4$, $y^2 = 4(4) = 16 \implies y = \pm 4$. The boundaries along the y-axis span from $y = -4$ to $y = 4$. Express the inner function as $x = \frac{y^2}{4}$. The horizontal strip extends from the curve on the left to the line $x = 4$ on the right: \[ A = \int_{-4}^4 \left( 4 - \frac{y^2}{4} \right) \, dy \] Since the integrand is an even function, we can simplify the boundaries: \[ A = 2 \int_0^4 \left( 4 - \frac{y^2}{4} \right) \, dy = 2 \left[ 4y - \frac{y^3}{12} \right]_0^4 \] \[ A = 2 \left[ 4(4) - \frac{4^3}{12} \right] = 2 \left[ 16 - \frac{64}{12} \right] = 2 \left[ 16 - \frac{16}{3} \right] = 2 \left[ \frac{32}{3} \right] = \frac{64}{3} \]
Both slicing orientations match precisely. Final Answer: $\frac{64}{3}$ square units.
✎ Self-Check — 5 questions0 / 5
Q1.Find the area enclosed between the curve $y = \ln x$, the x-axis, and the vertical bounding line $x = e$:
Q2.The area bounded by the curve $y = \cos x$ and the x-axis between the interval boundaries $x = 0$ and $x = \pi$ is:
Q3.Evaluate the area bounded between the curve $x = 4 - y^2$ and the vertical y-axis trace line:
Q4.The absolute physical area enclosed by the cubic function $y = x^3$ and the x-axis over the symmetric interval $[-1, 1]$ is:
Q5.Find the area of the region bounded by the curve $y = e^x$, the line $x = 0$, and the horizontal line $y = e$ using horizontal strips:

Area Between Two Intersecting Curves and Parametric ProfilesTopic 2

To evaluate the area bounded between two intersecting continuous curves $y = f(x)$ and $y = g(x)$ over an interval $[a, b]$, you must first determine their exact points of intersection by solving the simultaneous system $f(x) = g(x)$. The area is given by the integral of the absolute difference between the functions: \[ A = \int_a^b |f(x) - g(x)| \, dx \] If $f(x) \ge g(x)$ across the entire interval, this simplifies to $A = \int_a^b (f(x) - g(x)) \, dx$. If the curves cross each other at an interior point $c \in (a, b)$, you must split the integral into sub-intervals to ensure you are always subtracting the lower function from the upper function.

For curves defined implicitly by parametric equations $x = x(t)$ and $y = y(t)$ as the parameter varies from $t_1$ to $t_2$, the area mapping formula is transformed using differentials ($dx = \dot{x}(t) \, dt$): \[ A = \int_{t_1}^{t_2} |y(t) \cdot \dot{x}(t)| \, dt \quad \text{where } \dot{x}(t) = \frac{dx}{dt} \]

Worked Examples
1

Find the area of the region enclosed between the two standard parabolas $y^2 = x$ and $x^2 = y$.

Show solution

First, determine the points of intersection by substituting $y = x^2$ into the first parabola equation: \[ (x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0 \implies x = 0 \quad \text{and} \quad x = 1 \] The corresponding intersection coordinates are $(0, 0)$ and $(1, 1)$. Within the interval $[0, 1]$, notice that the curve $y = \sqrt{x}$ lies above the parabola $y = x^2$. Set up the area integral by subtracting the lower curve from the upper curve: \[ A = \int_0^1 \left( \sqrt{x} - x^2 \right) \, dx \] Integrate both components using the standard power rule: \[ A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_0^1 = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 \right]_0^1 \] Evaluate across the boundaries: \[ A = \left( \frac{2}{3}(1) - \frac{1}{3}(1) \right) - 0 = \frac{1}{3} \] Final Answer: $\frac{1}{3}$ square units.

✎ Self-Check — 5 questions0 / 5
Q1.Find the area enclosed between the parabola $y = x^2$ and the linear line $y = x$:
Q2.Find the area bounded by the curves $y = \sin x$ and $y = \cos x$ between two consecutive intersection points spanning from $x = \pi/4$ to $x = 5\pi/4$:
Q3.The area enclosed by the classical ellipse profile given parametrically as $x = a\cos t$ and $y = b\sin t$ ($t \in [0, 2\pi]$) is:
Q4.Evaluate the area bounded between the parabola $y^2 = 4x$ and the linear line $y = 2x$:
Q5.Find the total area enclosed between the line $y = x - 1$ and the parabola $y^2 = 2x + 6$:
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Module 2

Solid Geometry & Volumes of Revolution

Disk, Washer, and Cylindrical Shell Integration MethodsTopic 1

When a planar region bounded by continuous curves is rotated $360^\circ$ around a coordinate axis, it generates a three-dimensional solid of revolution. The volume of this solid can be calculated using three standard integration methods, chosen based on the axis of rotation and the orientation of the slicing boundaries:
  1. Disk Method: Used when the planar region is bounded by a single curve $y = f(x)$ and is rotated directly around the adjacent x-axis boundary. The solid is sliced into circular cross-sectional disks of radius $R = f(x)$ and thickness $dx$: \[ V = \pi \int_a^b [f(x)]^2 \, dx \]
  2. Washer Method: Used when the region bounded between two nested curves $y = f(x)$ (outer curve) and $y = g(x)$ (inner curve) is rotated around the x-axis. This creates a circular cross-section with a central hole, resembling a washer. The cross-sectional area is the difference between the outer and inner circular areas: \[ V = \pi \int_a^b \left[ [f(x)]^2 - [g(x)]^2 \right] \, dx \]
  3. Cylindrical Shell Method: An effective alternative when rotating a region bounded by a function $y = f(x)$ around the *opposite* vertical y-axis. Instead of cross-sectional disks, the region is integrated as a nested set of thin cylindrical shells of radius $x$, height $f(x)$, and thickness $dx$: \[ V = 2\pi \int_a^b x \cdot f(x) \, dx \]
Worked Examples
1

Find the volume of the solid generated by rotating the region bounded by $y = x^2$, the x-axis, and the vertical lines $x = 1$ and $x = 2$ around the x-axis.

Show solution

The region is bounded by a single curve and rotated directly around the x-axis, so we use the Disk Method: \[ V = \pi \int_a^b [f(x)]^2 \, dx \] Substitute the function $f(x) = x^2$ and the interval boundaries $a = 1, b = 2$ into the formula: \[ V = \pi \int_1^2 (x^2)^2 \, dx = \pi \int_1^2 x^4 \, dx \] Integrate using the standard power rule: \[ V = \pi \left[ \frac{x^5}{5} \right]_1^2 = \frac{\pi}{5} \left[ 2^5 - 1^5 \right] = \frac{\pi}{5} [32 - 1] = \frac{31\pi}{5} \] Final Answer: $\frac{31\pi}{5}$ cubic units.

✎ Self-Check — 5 questions0 / 5
Q1.Find the volume of the solid generated by rotating the linear region $y = 2x$ between $x = 0$ and $x = 3$ around the x-axis using the Disk Method:
Q2.Use the Washer Method to find the volume of the solid generated by rotating the region bounded by $y = x$ and $y = x^2$ around the x-axis:
Q3.Find the volume of the solid generated by rotating the region bounded by $y = \sqrt{x}$ from $x = 0$ to $x = 4$ around the vertical y-axis using the Cylindrical Shell Method:
Q4.Rotating a standard circular boundary region $x^2 + y^2 = R^2$ completely around its central axis generates a solid sphere. Its volume can be derived as:
Q5.Find the volume of the solid generated by rotating the region bounded by $y = e^x$ from $x = 0$ to $x = 1$ around the x-axis:
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Module 3

Differential Metrics & Rectification Profiles

Arc Length Rectification in Cartesian and Parametric FormsTopic 1

Rectification is the process of calculating the exact arc length of a curved path using calculus. The length $L$ of a smooth, continuously differentiable curve can be calculated by integrating a differential arc length element $ds = \sqrt{dx^2 + dy^2}$ along the path. This differential element can be configured into two standard forms depending on how the curve is defined:
  • Cartesian Form: For a curve defined explicitly as $y = f(x)$ over the interval $[a, b]$, factor out $dx$ from the differential element to express the arc length as: \[ L = \int_a^b \sqrt{1 + \left[ f'(x) \right]^2} \, dx = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
  • Parametric Form: For a curve defined implicitly by parametric coordinates $x = x(t)$ and $y = y(t)$ as the parameter varies from $t_1$ to $t_2$, factor out $dt$ to express the arc length as: \[ L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \int_{t_1}^{t_2} \sqrt{\dot{x}^2 + \dot{y}^2} \, dt \]
When solving these problems, look out for algebraic simplifications. Integrands containing square roots often simplify perfectly if you look for hidden perfect square structures ($\sqrt{1 + (f')^2} \to \sqrt{(u)^2} \to u$) before attempting to integrate.
Worked Examples
1

Find the arc length of the parametric circle profile given by $x = a\cos t$ and $y = a\sin t$ as the parameter tracks from $t = 0$ to $t = 2\pi$.

Show solution

Compute the individual rates of change with respect to the parameter $t$: \[ \dot{x} = \frac{dx}{dt} = -a\sin t, \quad \dot{y} = \frac{dy}{dt} = a\cos t \] Substitute these derivatives into the parametric arc length formula: \[ L = \int_0^{2\pi} \sqrt{(-a\sin t)^2 + (a\cos t)^2} \, dt \] Simplify the terms inside the radical using the standard Pythagorean trigonometric identity ($\sin^2 t + \cos^2 t = 1$): \[ L = \int_0^{2\pi} \sqrt{a^2\sin^2 t + a^2\cos^2 t} \, dt = \int_0^{2\pi} \sqrt{a^2(\sin^2 t + \cos^2 t)} \, dt = \int_0^{2\pi} a \, dt \] Integrate this constant form and evaluate across the boundaries to find the perimeter circumference: \[ L = a \Big[ t \Big]_0^{2\pi} = a(2\pi - 0) = 2\pi a \] Final Answer: $2\pi a$ units.

✎ Self-Check — 5 questions0 / 5
Q1.Find the exact arc length of the linear path segment $y = \frac{4}{3}x$ evaluated from $x = 0$ to $x = 3$:
Q2.Find the arc length of the parametric curve defined by $x = t^2$ and $y = \frac{2}{3}t^3$ from $t = 0$ to $t = 1$:
Q3.Which of the following equations precisely represents the integrand required to find the arc length of the curve $y = \frac{2}{3}x^{3/2}$?
Q4.Compute the perimeter arc length of the semicircular path $y = \sqrt{R^2 - x^2}$ from $x = -R$ to $x = R$:
Q5.Find the total length of the loop of the parametric curve $x = t^3 - 3t$ and $y = 3t^2$ from $t = -\sqrt{3}$ to $t = \sqrt{3}$:

Surface Area of Solids of RevolutionTopic 2

When a smooth curve $y = f(x)$ is rotated $360^\circ$ around a coordinate axis, the area of the resulting outer skin is defined as its surface area of revolution. To calculate this surface area, integrate the circumference of a differential circular ring element ($2\pi \cdot \text{radius}$) scaled by the differential arc length element $ds$ along the path.

For a continuously differentiable curve $y = f(x)$ bounded over the interval $[a, b]$ and rotated around the horizontal x-axis, the radius of rotation at any point is its vertical distance $y = f(x)$. The surface area is given by the integral formula: \[ S = 2\pi \int_a^b f(x) \cdot ds = 2\pi \int_a^b f(x) \sqrt{1 + \left[ f'(x) \right]^2} \, dx \]

This formula provides an analytical method for deriving standard geometric surface areas (such as the lateral surface area of cones or spheres) by rotating basic linear or circular curves around a central axis.

Worked Examples
1

Derive the surface area of a sphere of radius $R$ by rotating the upper semicircular arc $y = \sqrt{R^2 - x^2}$ from $x = -R$ to $x = R$ around the x-axis.

Show solution

First, calculate the derivative of the semicircular function: \[ f'(x) = \frac{d}{dx}\left(\sqrt{R^2 - x^2}\right) = \frac{-2x}{2\sqrt{R^2 - x^2}} = -\frac{x}{\sqrt{R^2 - x^2}} \] Square the derivative expression and substitute it into the standard arc length element: \[ \left[ f'(x) \right]^2 = \frac{x^2}{R^2 - x^2} \implies 1 + \left[ f'(x) \right]^2 = 1 + \frac{x^2}{R^2 - x^2} = \frac{R^2 - x^2 + x^2}{R^2 - x^2} = \frac{R^2}{R^2 - x^2} \] Take the square root of this expression to find the differential element: \[ \sqrt{1 + \left[ f'(x) \right]^2} = \frac{R}{\sqrt{R^2 - x^2}} \] Substitute this element and the function $y = \sqrt{R^2 - x^2}$ into the surface area formula: \[ S = 2\pi \int_{-R}^R \left(\sqrt{R^2 - x^2}\right) \cdot \frac{R}{\sqrt{R^2 - x^2}} \, dx \] Notice that the radical terms cancel out perfectly, leaving a simple constant integrand: \[ S = 2\pi \int_{-R}^R R \, dx = 2\pi R \Big[ x \Big]_{-R}^R = 2\pi R \left[ R - (-R) \right] = 2\pi R [2R] = 4\pi R^2 \] Final Answer: $4\pi R^2$ square units.

✎ Self-Check — 5 questions0 / 5
Q1.Find the surface area generated by rotating the line segment $y = x$ from $x = 0$ to $x = 1$ around the horizontal x-axis:
Q2.The formula used to calculate the surface area of revolution generated by rotating a smooth curve around the horizontal x-axis is:
Q3.Find the surface area generated by rotating the curve $y = 2x$ from $x = 0$ to $x = 2$ around the x-axis:
Q4.Find the surface area of the solid generated by rotating the upper half of the loop of the curve $9y^2 = x(3-x)^2$ around the x-axis from $x = 0$ to $x = 3$:
Q5.If a vertical line segment of length $H$ located at a distance $R$ from the origin is rotated around the vertical y-axis, it generates a cylinder. Its lateral surface area can be derived as:

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