第 12 类 RD Sharma 解决方案 - 第 21 章有界区域 - 练习 21.2

问题 1. 找出第一象限中由抛物线 y = 4x ²和线 x = 0、y = 1和 y = 4 界定的区域。

解决方案：

From the question it is given that,

Lines, x = 0, y = 1, y = 4

Parabola y = 4x² … [equation (i)]

So, equation (i) represents a parabola with vertex (0, 0) and axis as y – axis. x = 0 is y – axis and y = 1, y = 4 are line parallel to x – axis passing through (0, 1) and (0, 4) respectively, as shown in the rough sketch below,

Now, we have to find the area of ABCDA,

Then, the area can be found by taking a small slice in each region of width Δy,

And length = x

The area of sliced part will be as it is a rectangle = x Δy

So, this rectangle can move horizontal from y = 1 to x = 4

The required area of the region bounded between the lines = Region ABCDA

$\displaystyle=\int_1^4xdy$

Given, y = 4x²

x = $\displaystyle\sqrt{\frac{y}{4}}\\ \int_1^4\sqrt{\frac{y}{4}}\ dy\\ \frac{1}{2}\int_1^4\sqrt y\ dy$

On integration, we get,

$\displaystyle=\frac{1}{2}\left[\frac{2}{3}y\sqrt y\right]_1^4$

Now, applying limits we get,

$= \frac{1}{2} \left[\left(\frac{2}{3} × 4 × \sqrt4\right) - \left(\frac{2}{3} × 1 × \sqrt1\right)\right]\\ = \frac{1}{2} \left[\frac{16}{3} - \frac{2}{3}\right]\\ = \frac{1}{2} \left[\frac{(16 - 2)}{3}\right]\\ = \frac{1}{2} \left[\frac{14}{3}\right]\\ = \frac{7}{3}$

Therefore, the required area is $\frac{7}{3}$ square units.

问题 2. 求象限中以 x 2 = 16y、y = 1、y = 4 和 y 轴为^界的区域的面积。

解决方案：

From the question it is given that,

Region in first quadrant bounded by y = 1, y = 4

Parabola x² = 16y … [equation (i)]

So, equation (i) represents a parabola with vertex (0, 0) and axis as y-axis, as shown in the rough sketch below,

Now, we have to find the area of ABCDA,

Then, the area can be found by taking a small slice in each region of width Δy,

And length = x

The area of sliced part will be as it is a rectangle = x Δy

So, this rectangle can move horizontal from y = 1 to x = 4

The required area of the region bounded between the lines = Region ABCDA

$\displaystyle\int_1^4x\ dy$

Given, x² = 16y

$x=\sqrt{16y}\\ x=4\sqrt{y}\\ =\int_1^44\sqrt4\ dy$ 

On integrating we get,

$=4\left[\frac{2}{3}y\sqrt y\right]^4_1\\ =\int_1^4x\ dy$

Given, x² = 16y

$x=\sqrt{16y}\\ x=4\sqrt{y}\\ =\int_1^44\sqrt4\ dy$

On integrating we get,

$=4\left[\frac{2}{3}y\sqrt y\right]^4_1\\ =\int_1^4x\ dy$

Now, applying limits we get,

$= 4 \left[\left(\frac{2}{3} × 4 × \sqrt4\right) - \left(\frac{2}{3}× 1 × \sqrt1\right)\right]\\ = 4 \left[\frac{16}{3} - \frac{2}{3}\right]\\ = 4 \left[\frac{(16 - 2)}{3}\right]\\ = 4 \left[\frac{14}{3}\right]\\ = \frac{56}{3}$

Therefore, the required area is $\frac{56}{3}$ square units.

问题 3. 求以 x ² = 4ay 及其直角为界的区域面积。

解决方案：

We have to find the area of the region bounded by x² = 4ay

Then,

Area of the region = $\displaystyle2\times\int_0^{2a}\left(a-\frac{x^2}{4a}\right)dx$

On integrating we get,

= $2\times\left[ax-\frac{x^3}{12a}\right]_0^{2a}$

Now applying limits,

$= 2 × \left[(a (2a - 0)) - \frac{((2a)^3 - 0^3)}{12a}\right]\\ = 2 × \left[(2a^2) - \frac{8a^3}{12a}\right]\\ = 2 × \left[\frac{(24a^3 - 8a^3)}{12a}\right]\\ = 2 × \left[\frac{16a^3}{12a}\right]\\ = 2 × \left[\frac{4a^2}{3}\right]\\ = \frac{8a^2}{3}\\$

Therefore, the area of the region is $\frac{8a^2}{3}$ square units.

问题 4. 求以 x ² + 16y = 0 及其直角为界的区域的面积。

解决方案：

We have to find the area of the region bounded by x² + 16y = 0

Then, 

Area of the region = $\displaystyle2\times\int_0^{8}\left[-\frac{x^2}{16}-(-4)\right]dx$

On integrating we get,

= $2\times\left[4x-\frac{x^3}{48}\right]_0^{8}$

Now applying limits,

$= 2 × \left[(4 (8 - 0)) - \frac{((8)^3 - 0^3)}{48}\right]\\ = 2 × \left[(32) - \frac{512}{48}\right]\\ = 2 × \left[(32)-\frac{32}{3}\right]\\ = 2 × \left[\frac{96-32}{3}\right]\\ = 2 × \left[\frac{64}{3}\right]\\ = \frac{128}{3}\\$

Therefore, the area of the region is $\frac{128}{3}$ square units.

问题 5. 求曲线 ay 2 = x 3 、y 轴和直线 y = a 和 y = 2a^所包围的区域的面积。

解决方案：

We have to find the area of the region bounded by curve ay² = x³, and lines y = a and y = 2a.

Then,

Area of the region = $\displaystyle\int_a^{2a}(ay^2)^{\frac{1}{3}}\ dy\\ =a^{\frac{1}{3}}\int_a^{2a}y^{\frac{2}{3}}dy$

On integrating we get, = $a^{\frac{1}{3}}\left[\frac{3}{5}y^{\frac{5}{3}}\right]_0^{2a}$

Now applying limits we get, = $\frac{3}{5}\left(2^{\frac{5}{3}}-1\right)a^2\ sq\ units$