Application of Integrals - Class 12 Mathematics - Chapter 8 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Application of Integrals | NCERT | Mathematics | Class 12
The area bounded by the curve $x=a \cos^3 t, y=a \sin^3 t$ is:
A) $\frac{3 \pi a^2}{8}$
B) $\frac{3 \pi a^2}{16}$
C) $\frac{3 \pi a^2}{32}$
D) $3 \pi a^2$
The question involves calculating the area enclosed by the curve given in parametric form as:
$$ x = a \cos^3 t, \quad y = a \sin^3 t $$
Step 1: Curve Equation Transformation
By eliminating the parameter $t$ between the equations of $x$ and $y$, we obtain: $$ x^{2/3} + y^{2/3} = a^{2/3} $$ This equation describes a shape symmetric about both axes.
Step 2: Calculating the Area
To find the enclosed area $A$, the general formula for the area under a curve in parametric form is: $$ A = \int y , dx $$ Given the symmetry, the full area is four times the area in the first quadrant: $$ A = 4 \int y , dx $$
Step 3: Parametric Derivatives and Integration
Using $ x = a \cos^3 t $ and $ y = a \sin^3 t $, $$ dx = -3a\cos^2 t \sin t , dt $$ Thus, the integral for the area becomes: $$ A = 4 \int_{0}^{\pi/2} (a \sin^3 t)(-3 a \cos^2 t \sin t) , dt $$ $$ A = 12a^2 \int_{0}^{\pi/2} \sin^4 t \cos^2 t , dt $$
Step 4: Simplifying the Integral
Using the symmetry and substitution $ t \rightarrow \frac{\pi}{2} - t $: $$ A = 12a^2 \int_{0}^{\pi/2} \cos^4 t \sin^2 t , dt $$ Combining both integrals due to symmetry of $\sin$ and $\cos$: $$ A = 6a^2 \int_{0}^{\pi/2} \sin^2 t \cos^2 t , dt $$ Using the double angle identity: $$ \sin^2 t \cos^2 t = \frac{1}{4} \sin^2 2t $$ The integral becomes: $$ A = \frac{3a^2}{2} \int_{0}^{\pi/2} \sin^2 2t , dt $$ Using another integral transformation for $ \sin^2 2t $, $$ A = \frac{3a^2}{2} \int_{0}^{\pi/4}(1 - \cos 4t) , dt $$ After evaluating the integral, $$ A = \frac{3\pi a^2}{8} $$
Final Answer: The area bounded by the curve is $\frac{3 \pi a^2}{8}$, matching option A) $\frac{3 \pi a^2}{8}$.
Choose the correct answer: The area bounded by the $Y$-axis, $y = \cos x$, and $y = \sin x$, $0 \leq x \leq \frac{\pi}{2}$ is (a) $2(\sqrt{2}-1)$ (b) $\sqrt{2}-1$ (c) $\sqrt{2}+1$ (d) $\sqrt{2}$
The goal is to find the area enclosed between the curves $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{2}$.
Intersection Point: First, we need to establish where these two curves intersect within the given range. This occurs where $\sin x = \cos x$. Solving this equation, using the identity for tangent, $$ \tan x = 1, \Rightarrow x = \frac{\pi}{4} $$
Setting Up the Integral: Each function dominates the other in different intervals from 0 to $\frac{\pi}{2}$. Over the interval $0$ to $\frac{\pi}{4}$, $\cos x$ is greater than $\sin x$. Hence, to find the area, an integral of their difference, $\cos x - \sin x$, over this interval gets the required area: $$ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) , dx $$
Evaluating the Integral: This integral evaluates as follows: $$ \left[ \sin x + \cos x \right]_{0}^{\frac{\pi}{4}} = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right) $$ Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin 0 = 0$, $\cos 0 = 1$, the above expression simplifies to: $$ \sin \frac{\pi}{4} + \cos \frac{\pi}{4} - 1 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = \sqrt{2} - 1 $$
Conclusion: Therefore, the correct answer is (b) $\sqrt{2} - 1$.
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