Area of Standard Curves

Area of Standard Curves: Your JEE Main Ticket

Hello students! Welcome to a crucial topic for your JEE Main exam: Area of Standard Curves. This isn't just about memorizing formulas; it's about understanding how calculus lets us calculate areas bounded by familiar shapes. Mastering this will not only boost your score but also deepen your understanding of integration. So, let's dive in!

The Conceptual Foundation

At its core, finding the area under a curve involves integration. Imagine slicing the area into infinitely thin rectangles. The sum of the areas of these rectangles, as their width approaches zero, gives us the exact area. This is the essence of definite integration.

1. Area of a Circle Using Integration

Let's start with the most basic shape: the circle. Consider a circle with radius $r$ centered at the origin, represented by the equation $x^2 + y^2 = r^2$.

Derivation:

1. Solve for $y$: $y = \pm \sqrt{r^2 - x^2}$. The positive root represents the upper half of the circle, and the negative root represents the lower half.
2. To find the total area, we'll calculate the area of the upper half and double it: $$Area = 2 \int_{-r}^{r} \sqrt{r^2 - x^2} \, dx$$
3. To solve this integral, use the trigonometric substitution $x = r \sin(\theta)$, so $dx = r \cos(\theta) \, d\theta$. The limits of integration change accordingly: when $x = -r$, $\theta = -\frac{\pi}{2}$, and when $x = r$, $\theta = \frac{\pi}{2}$.
4. The integral becomes: $$Area = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^2 - r^2 \sin^2(\theta)} \cdot r \cos(\theta) \, d\theta$$ $$Area = 2r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta$$
5. Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$: $$Area = 2r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta$$ $$Area = r^2 \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$ $$Area = r^2 \left[ \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right] = \pi r^2$$

Thus, the area of the circle is indeed $\pi r^2$.

2. Area of an Ellipse

Now, let's move to the ellipse. An ellipse is like a stretched circle, with two axes: a major axis ($2a$) and a minor axis ($2b$). The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Derivation:

1. Solve for $y$: $y = \pm \frac{b}{a} \sqrt{a^2 - x^2}$. Again, we take the positive root for the upper half.
2. The area of the ellipse is twice the integral of the upper half: $$Area = 2 \int_{-a}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx$$
3. Notice that $\frac{b}{a}$ is a constant, so we can pull it out: $$Area = \frac{2b}{a} \int_{-a}^{a} \sqrt{a^2 - x^2} \, dx$$
4. The integral is the same as the one we solved for the circle, except with radius $a$. So, $\int_{-a}^{a} \sqrt{a^2 - x^2} \, dx = \frac{\pi a^2}{2}$.
5. Substitute back: $$Area = \frac{2b}{a} \cdot \frac{\pi a^2}{2} = \pi ab$$

Therefore, the area of the ellipse is $\pi ab$. Notice that if $a = b = r$, we get the area of a circle, $\pi r^2$, which makes sense!

3. Area Bounded by a Parabola

Parabolas are also common in JEE problems. Let's consider the standard parabola $y^2 = 4ax$. The latus rectum is a line segment through the focus, perpendicular to the axis of symmetry, with endpoints on the parabola.

Derivation:

1. The latus rectum of $y^2 = 4ax$ is the line $x = a$. The parabola intersects the latus rectum at points $(a, 2a)$ and $(a, -2a)$.
2. We need to find the area bounded by the parabola and the line $x = a$. $$Area = \int_{-2a}^{2a} (a - \frac{y^2}{4a}) \, dy$$
3. Since the parabola is symmetric about the x-axis, we can calculate the area in the first quadrant and double it: $$Area = 2 \int_{0}^{2a} (a - \frac{y^2}{4a}) \, dy$$
4. Integrate: $$Area = 2 \left[ ay - \frac{y^3}{12a} \right]_{0}^{2a}$$ $$Area = 2 \left[ a(2a) - \frac{(2a)^3}{12a} \right] = 2 \left[ 2a^2 - \frac{8a^3}{12a} \right] = 2 \left[ 2a^2 - \frac{2a^2}{3} \right] = 2 \cdot \frac{4a^2}{3} = \frac{8a^2}{3}$$

Thus, the area bounded by the parabola and its latus rectum is $\frac{8a^2}{3}$.

4. Area of a Parabola Segment

A parabolic segment is the region bounded by a parabola and a chord. The area of a parabola segment can be found using integration, but there's a handy shortcut formula if you know the base and height of the segment. Unfortunately, a general formula is complicated and depends on the exact parabola and chord. If the chord is the latus rectum (as discussed above) we already have the formula! For other chords, you'll need to integrate.

Example: Let's find the area of the region bounded by the parabola $y = x^2$ and the line $y = 4$.

1. Find the intersection points: $x^2 = 4$, so $x = \pm 2$. The points are $(-2, 4)$ and $(2, 4)$.
2. Set up the integral: $$Area = \int_{-2}^{2} (4 - x^2) \, dx$$
3. Integrate: $$Area = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = 16 - \frac{16}{3} = \frac{32}{3}$$

Therefore, the area of this parabolic segment is $\frac{32}{3}$.

JEE-Specific Tricks

• Quick Recognition: Practice recognizing standard curves and their properties quickly. Time is precious in JEE!
• Standard Integrals: Memorize the integrals of common functions like $\sqrt{a^2 - x^2}$, $\sin^2(x)$, and $\cos^2(x)$.
• Symmetry: Always check for symmetry. It can drastically reduce the amount of calculation needed.

That's it for this lesson! Keep practicing, and you'll master the area of standard curves. All the best for your JEE Main preparation!