Application of IntegralsClass 12 Mathematics NCERT Solutions

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get: $a^2 = 16 \implies a = 4$ $b^2 = 9 \implies b = 3$

The ellipse is symmetrical about both the x-axis and the y-axis. Therefore, the total area of the ellipse is 4 times the area of the region in the first quadrant.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16 - x^2}{16}$ $y^2 = \frac{9}{16}(16 - x^2)$ $y = \pm \frac{3}{4}\sqrt{16 - x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{4}\sqrt{16 - x^2}$. The area of the region in the first quadrant is bounded by $x=0$ and $x=a=4$.

Area of ellipse, $A = 4 \times (\text{Area in the first quadrant})$ $A = 4 \int_{0}^{4} y \, dx$ $A = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx$ $A = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$. Here, $a=4$. $A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_0^4$ $A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4} \right]_0^4$ $A = 3 \left( \left( \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\frac{4}{4} \right) - \left( \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\frac{0}{4} \right) \right)$ $A = 3 \left( (2\sqrt{0} + 8\sin^{-1}(1)) - (0 + 8\sin^{-1}(0)) \right)$ $A = 3 \left( (0 + 8 \times \frac{\pi}{2}) - (0 + 0) \right)$ $A = 3 (4\pi) = 12\pi$

Alternatively, using the direct formula for the area of an ellipse, Area $= \pi ab = \pi(4)(3) = 12\pi$.

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{4^2}+\frac{y^{2}}{3^2}=1$. This is a standard ellipse with semi-major axis $a=4$ and semi-minor axis $b=3$. The ellipse is symmetric about both the x-axis and y-axis.

We can find the area of the region in the first quadrant and multiply it by 4 to get the total area of the ellipse.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16-x^2}{16}$ $y^2 = \frac{9}{16}(16-x^2)$ $y = \pm \frac{3}{4}\sqrt{16-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{4}\sqrt{16-x^2}$. The limits of integration for the first quadrant are from $x=0$ to $x=a=4$.

Area of the ellipse = $4 \times (\text{Area in the first quadrant})$ $A = 4 \int_{0}^{4} y \, dx$ $A = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \, dx$ $A = 3 \int_{0}^{4} \sqrt{4^2-x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$. Here, $a=4$.

$A = 3 \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$ $A = 3 \left[ \left( \frac{4}{2}\sqrt{16-4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) - \left( \frac{0}{2}\sqrt{16-0^2} + 8\sin^{-1}\left(\frac{0}{4}\right) \right) \right]$ $A = 3 \left[ (2\sqrt{0} + 8\sin^{-1}(1)) - (0 + 8\sin^{-1}(0)) \right]$ $A = 3 \left[ (0 + 8 \times \frac{\pi}{2}) - (0 + 0) \right]$ $A = 3 [4\pi] = 12\pi$

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse can be written as $\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$. This is a standard ellipse with semi-major axis $a=4$ and semi-minor axis $b=3$.

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16-x^2}{16}$ $y^2 = \frac{9}{16}(16-x^2)$ $y = \pm \frac{3}{4}\sqrt{16-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{4}\sqrt{16-x^2}$. The area of the ellipse in the first quadrant is bounded by the curve, the x-axis, and the ordinates $x=0$ and $x=4$.

Area of region in first quadrant = $\int_{0}^{4} y \,dx = \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \,dx$

Total area of the ellipse = $4 \times (\text{Area of region in first quadrant})$ $A = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \,dx = 3 \int_{0}^{4} \sqrt{4^2-x^2} \,dx$

Using the formula $\int \sqrt{a^2-x^2} \,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$, we get: $A = 3 \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_{0}^{4}$ $A = 3 \left[ \left(\frac{4}{2}\sqrt{16-4^2} + 8\sin^{-1}\frac{4}{4}\right) - \left(\frac{0}{2}\sqrt{16-0^2} + 8\sin^{-1}\frac{0}{4}\right) \right]$ $A = 3 \left[ (2\sqrt{0} + 8\sin^{-1}(1)) - (0 + 8\sin^{-1}(0)) \right]$ $A = 3 \left[ (0 + 8 \times \frac{\pi}{2}) - 0 \right]$ $A = 3 (4\pi) = 12\pi$

Alternatively, using the direct formula for the area of an ellipse, Area = $\pi ab$. Here $a=4$ and $b=3$. So, Area = $\pi(4)(3) = 12\pi$.

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{4^2} + \frac{y^{2}}{3^2} = 1$. This is a standard ellipse with semi-major axis $a=4$ and semi-minor axis $b=3$.

The ellipse is symmetrical about both the x-axis and y-axis. Therefore, the total area of the ellipse is 4 times the area of the region in the first quadrant.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16}$ $y^2 = 9(1 - \frac{x^2}{16}) = \frac{9}{16}(16 - x^2)$ $y = \pm \frac{3}{4}\sqrt{16 - x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{4}\sqrt{16 - x^2}$. The area of the region in the first quadrant is bounded by $x=0$ and $x=a=4$.

Area of ellipse $= 4 \times (\text{Area in the first quadrant})$ $A = 4 \int_{0}^{4} y \, dx$ $A = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx$ $A = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$. Here, $a=4$. $A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_{0}^{4}$ $A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4} \right]_{0}^{4}$

Substitute the upper and lower limits: $A = 3 \left[ \left( \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\frac{4}{4} \right) - \left( \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\frac{0}{4} \right) \right]$ $A = 3 \left[ (2\sqrt{0} + 8\sin^{-1}(1)) - (0 + 8\sin^{-1}(0)) \right]$ $A = 3 \left[ (0 + 8 \times \frac{\pi}{2}) - (0 + 0) \right]$ $A = 3 [4\pi] = 12\pi$

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. Comparing this with the standard equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get $a^2 = 16$ and $b^2 = 9$. Thus, $a=4$ and $b=3$.

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16 - x^2}{16}$ $y^2 = \frac{9}{16}(16 - x^2)$ $y = \frac{3}{4}\sqrt{16 - x^2}$ In the first quadrant, $y$ is positive. The limits of integration for the first quadrant are from $x=0$ to $x=a=4$.

Area of the ellipse = $4 \times$ (Area of the region in the first quadrant) $A = 4 \int_{0}^{4} y \, dx$ $A = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx$ $A = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

We use the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

$A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$ $A = 3 \left[ \left( \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) - \left( \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\left(\frac{0}{4}\right) \right) \right]$ $A = 3 \left[ \left( 2\sqrt{0} + 8\sin^{-1}(1) \right) - (0 + 8\sin^{-1}(0)) \right]$ $A = 3 \left[ (0 + 8 \times \frac{\pi}{2}) - 0 \right]$ $A = 3 [4\pi]$ $A = 12\pi$

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{4^2} + \frac{y^2}{3^2} = 1$. This is a standard ellipse with semi-major axis $a=4$ and semi-minor axis $b=3$.

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16-x^2}{16}$ $y^2 = \frac{9}{16}(16-x^2)$ $y = \pm \frac{3}{4}\sqrt{16-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{4}\sqrt{16-x^2}$. The area of the ellipse in the first quadrant is bounded by $x=0$ and $x=4$.

Area of ellipse in the first quadrant = $\int_{0}^{4} y dx$ $= \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} dx = \frac{3}{4} \int_{0}^{4} \sqrt{4^2-x^2} dx$

We use the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$. Here, $a=4$.

$= \frac{3}{4} \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_{0}^{4}$ $= \frac{3}{4} \left[ \left( \frac{4}{2}\sqrt{16-4^2} + 8\sin^{-1}\frac{4}{4} \right) - \left( \frac{0}{2}\sqrt{16-0^2} + 8\sin^{-1}\frac{0}{4} \right) \right]$ $= \frac{3}{4} \left[ (2 \times 0 + 8\sin^{-1}1) - (0 + 8\sin^{-1}0) \right]$ $= \frac{3}{4} \left[ 8 \times \frac{\pi}{2} - 0 \right] = \frac{3}{4} \times 4\pi = 3\pi$

Total area of the ellipse = $4 \times$ (Area in the first quadrant) $= 4 \times 3\pi = 12\pi \text{ square units}$

Final Answer: The area of the region bounded by the ellipse is $12\pi$ square units.

Given: The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse can be written as $\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1$. This is a standard ellipse with semi-major axis $b=3$ (along the y-axis) and semi-minor axis $a=2$ (along the x-axis).

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{4} = \frac{4-x^2}{4}$ $y^2 = \frac{9}{4}(4-x^2)$ $y = \pm \frac{3}{2}\sqrt{4-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{2}\sqrt{4-x^2}$. The area of the ellipse in the first quadrant is bounded by the curve, the x-axis, and the ordinates $x=0$ and $x=2$.

Area of region in first quadrant = $\int_{0}^{2} y \,dx = \int_{0}^{2} \frac{3}{2}\sqrt{4-x^2} \,dx$

Total area of the ellipse = $4 \times (\text{Area of region in first quadrant})$ $A = 4 \int_{0}^{2} \frac{3}{2}\sqrt{4-x^2} \,dx = 6 \int_{0}^{2} \sqrt{2^2-x^2} \,dx$

Using the formula $\int \sqrt{a^2-x^2} \,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$, we get: $A = 6 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2}$ $A = 6 \left[ \left(\frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\frac{2}{2}\right) - \left(\frac{0}{2}\sqrt{4-0^2} + 2\sin^{-1}\frac{0}{2}\right) \right]$ $A = 6 \left[ (1\sqrt{0} + 2\sin^{-1}(1)) - (0 + 2\sin^{-1}(0)) \right]$ $A = 6 \left[ (0 + 2 \times \frac{\pi}{2}) - 0 \right]$ $A = 6 (\pi) = 6\pi$

Alternatively, using the direct formula for the area of an ellipse, Area = $\pi ab$. Here $a=2$ and $b=3$. So, Area = $\pi(2)(3) = 6\pi$.

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{2^2}+\frac{y^{2}}{3^2}=1$. This is a standard ellipse with semi-major axis $b=3$ along the y-axis and semi-minor axis $a=2$ along the x-axis. The ellipse is symmetric about both the x-axis and y-axis.

We can find the area of the region in the first quadrant and multiply it by 4 to get the total area of the ellipse.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{4} = \frac{4-x^2}{4}$ $y^2 = \frac{9}{4}(4-x^2)$ $y = \pm \frac{3}{2}\sqrt{4-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{2}\sqrt{4-x^2}$. The limits of integration for the first quadrant are from $x=0$ to $x=a=2$.

Area of the ellipse = $4 \times (\text{Area in the first quadrant})$ $A = 4 \int_{0}^{2} y \, dx$ $A = 4 \int_{0}^{2} \frac{3}{2}\sqrt{4-x^2} \, dx$ $A = 6 \int_{0}^{2} \sqrt{2^2-x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$. Here, $a=2$.

$A = 6 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$ $A = 6 \left[ \left( \frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4-0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right) \right]$ $A = 6 \left[ (1\sqrt{0} + 2\sin^{-1}(1)) - (0 + 2\sin^{-1}(0)) \right]$ $A = 6 \left[ (0 + 2 \times \frac{\pi}{2}) - (0 + 0) \right]$ $A = 6 [\pi] = 6\pi$

Alternatively, for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the area is $\pi ab$. Here $a=2$ and $b=3$, so Area = $\pi(2)(3) = 6\pi$.

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. Comparing this with the standard equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get $a^2 = 4$ and $b^2 = 9$. Thus, $a=2$ and $b=3$.

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{4} = \frac{4 - x^2}{4}$ $y^2 = \frac{9}{4}(4 - x^2)$ $y = \frac{3}{2}\sqrt{4 - x^2}$ In the first quadrant, $y$ is positive. The limits of integration for the first quadrant are from $x=0$ to $x=a=2$.

Area of the ellipse = $4 \times$ (Area of the region in the first quadrant) $A = 4 \int_{0}^{2} y \, dx$ $A = 4 \int_{0}^{2} \frac{3}{2}\sqrt{4 - x^2} \, dx$ $A = 6 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$

We use the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

$A = 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$ $A = 6 \left[ \left( \frac{2}{2}\sqrt{4 - 2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right) \right]$ $A = 6 \left[ \left( 1\sqrt{0} + 2\sin^{-1}(1) \right) - (0 + 2\sin^{-1}(0)) \right]$ $A = 6 \left[ (0 + 2 \times \frac{\pi}{2}) - 0 \right]$ $A = 6 [\pi]$ $A = 6\pi$

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get: $a^2 = 4 \implies a = 2$ $b^2 = 9 \implies b = 3$

The area of an ellipse is given by the formula $A = \pi ab$. $A = \pi (2)(3) = 6\pi$

To solve using integration, we can integrate with respect to the y-axis since the major axis is along the y-axis ($b>a$). The ellipse is symmetrical about both axes. Area $= 4 \times (\text{Area in the first quadrant})$ From the equation, we express $x$ in terms of $y$: $\frac{x^2}{4} = 1 - \frac{y^2}{9} = \frac{9-y^2}{9}$ $x^2 = \frac{4}{9}(9-y^2)$ $x = \frac{2}{3}\sqrt{9-y^2}$ (in the first quadrant).

Area of ellipse, $A = 4 \int_{0}^{3} x \, dy$ $A = 4 \int_{0}^{3} \frac{2}{3}\sqrt{9-y^2} \, dy$ $A = \frac{8}{3} \int_{0}^{3} \sqrt{3^2-y^2} \, dy$ Using the standard integral formula: $\int \sqrt{a^2 - y^2} \, dy = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\frac{y}{a}$. $A = \frac{8}{3} \left[ \frac{y}{2}\sqrt{9-y^2} + \frac{9}{2}\sin^{-1}\frac{y}{3} \right]_0^3$ $A = \frac{8}{3} \left( \left( \frac{3}{2}\sqrt{9-9} + \frac{9}{2}\sin^{-1}\frac{3}{3} \right) - (0) \right)$ $A = \frac{8}{3} \left( 0 + \frac{9}{2}\sin^{-1}(1) \right) = \frac{8}{3} \left( \frac{9}{2} \times \frac{\pi}{2} \right)$ $A = \frac{8}{3} \times \frac{9\pi}{4} = 2 \times 3\pi = 6\pi$

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{2^2} + \frac{y^{2}}{3^2} = 1$. This is a standard ellipse with semi-major axis $b=3$ along the y-axis and semi-minor axis $a=2$ along the x-axis.

The ellipse is symmetrical about both the x-axis and y-axis. Therefore, the total area of the ellipse is 4 times the area of the region in the first quadrant.

It is more convenient to integrate with respect to $y$ since the major axis is along the y-axis. We express $x$ in terms of $y$: $\frac{x^2}{4} = 1 - \frac{y^2}{9}$ $x^2 = 4(1 - \frac{y^2}{9}) = \frac{4}{9}(9 - y^2)$ $x = \pm \frac{2}{3}\sqrt{9 - y^2}$

In the first quadrant, $x$ is positive, so we take $x = \frac{2}{3}\sqrt{9 - y^2}$. The area of the region in the first quadrant is bounded by $y=0$ and $y=b=3$.

Area of ellipse $= 4 \times (\text{Area in the first quadrant})$ $A = 4 \int_{0}^{3} x \, dy$ $A = 4 \int_{0}^{3} \frac{2}{3}\sqrt{9 - y^2} \, dy$ $A = \frac{8}{3} \int_{0}^{3} \sqrt{3^2 - y^2} \, dy$

We use the standard integral formula: $\int \sqrt{a^2-y^2} \, dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\frac{y}{a}$. Here, $a=3$. $A = \frac{8}{3} \left[ \frac{y}{2}\sqrt{9 - y^2} + \frac{9}{2}\sin^{-1}\frac{y}{3} \right]_{0}^{3}$

Substitute the upper and lower limits: $A = \frac{8}{3} \left[ \left( \frac{3}{2}\sqrt{9 - 3^2} + \frac{9}{2}\sin^{-1}\frac{3}{3} \right) - \left( \frac{0}{2}\sqrt{9 - 0^2} + \frac{9}{2}\sin^{-1}\frac{0}{3} \right) \right]$ $A = \frac{8}{3} \left[ (\frac{3}{2}\sqrt{0} + \frac{9}{2}\sin^{-1}(1)) - (0 + \frac{9}{2}\sin^{-1}(0)) \right]$ $A = \frac{8}{3} \left[ (0 + \frac{9}{2} \times \frac{\pi}{2}) - (0 + 0) \right]$ $A = \frac{8}{3} \left[ \frac{9\pi}{4} \right] = 2 \times 3\pi = 6\pi$

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.

To Find: The area of the region bounded by the ellipse.

Solution: The given equation of the ellipse is $\frac{x^{2}}{2^2} + \frac{y^2}{3^2} = 1$. This is a standard ellipse with semi-major axis $b=3$ (along y-axis) and semi-minor axis $a=2$ (along x-axis).

The ellipse is symmetrical about both the x-axis and y-axis. We can find the area of the region in the first quadrant and multiply it by 4 to get the total area.

From the equation of the ellipse, we can express $y$ in terms of $x$: $\frac{y^2}{9} = 1 - \frac{x^2}{4} = \frac{4-x^2}{4}$ $y^2 = \frac{9}{4}(4-x^2)$ $y = \pm \frac{3}{2}\sqrt{4-x^2}$

In the first quadrant, $y$ is positive, so we take $y = \frac{3}{2}\sqrt{4-x^2}$. The area of the ellipse in the first quadrant is bounded by $x=0$ and $x=2$.

Area of ellipse in the first quadrant = $\int_{0}^{2} y dx$ $= \int_{0}^{2} \frac{3}{2}\sqrt{4-x^2} dx = \frac{3}{2} \int_{0}^{2} \sqrt{2^2-x^2} dx$

We use the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$. Here, $a=2$.

$= \frac{3}{2} \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2}$ $= \frac{3}{2} \left[ \left( \frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\frac{2}{2} \right) - \left( \frac{0}{2}\sqrt{4-0^2} + 2\sin^{-1}\frac{0}{2} \right) \right]$ $= \frac{3}{2} \left[ (1 \times 0 + 2\sin^{-1}1) - (0 + 2\sin^{-1}0) \right]$ $= \frac{3}{2} \left[ 2 \times \frac{\pi}{2} - 0 \right] = \frac{3}{2} \times \pi = \frac{3\pi}{2}$

Total area of the ellipse = $4 \times$ (Area in the first quadrant) $= 4 \times \frac{3\pi}{2} = 6\pi \text{ square units}$

Final Answer: The area of the region bounded by the ellipse is $6\pi$ square units.

Given: The equation of the circle is $x^{2}+y^{2}=4$. The region is in the first quadrant, bounded by the lines $x=0$ and $x=2$.

To Find: The area of the specified region.

Solution: The equation of the circle can be written as $x^2 + y^2 = 2^2$. This represents a circle centered at the origin with a radius of $r=2$. The region is in the first quadrant bounded by the y-axis ($x=0$) and the line $x=2$. This describes the area of the circle in the first quadrant.

From the equation of the circle, we express $y$ in terms of $x$: $y^2 = 4 - x^2$ $y = \pm \sqrt{4 - x^2}$

Since the region is in the first quadrant, $y$ is positive. So, $y = \sqrt{4 - x^2}$.

The required area is given by the integral of $y$ from $x=0$ to $x=2$. $A = \int_{0}^{2} y \, dx = \int_{0}^{2} \sqrt{4-x^2} \, dx = \int_{0}^{2} \sqrt{2^2-x^2} \, dx$

Using the formula $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$ with $a=2$: $A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$ $A = \left[ \frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right] - \left[ \frac{0}{2}\sqrt{4-0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right]$ $A = \left[ 1\sqrt{0} + 2\sin^{-1}(1) \right] - [0 + 2\sin^{-1}(0)]$ $A = [0 + 2(\frac{\pi}{2})] - [0 + 0]$ $A = \pi$

This is expected, as the area is one-quarter of the total area of a circle with radius 2. Total area is $\pi r^2 = \pi (2^2) = 4\pi$. Quarter area is $\frac{4\pi}{4} = \pi$.

Answer: (A)

Explanation: The area required is the area of a quarter-circle of radius 2. The area of a full circle is $\pi r^2 = 4\pi$. The area of the quarter-circle is $\frac{1}{4}(4\pi) = \pi$ square units.