Question

solvw these questions ​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^a\int\limits_0^{\sqrt{a^2-y^2}}\sqrt{a^2-x^2-y^2}\ dx\ dy$

or,

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\int\limits_0^{\sqrt{a^2-y^2}}\sqrt{(a^2-y^2)-x^2}\ dx\right]\ dy$

Let $\sqrt{a^2-y^2}=c$ which is treated as a constant wrt $x.$ Then,

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\int\limits_0^c\sqrt{c^2-x^2}\ dx\right]\ dy$

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\dfrac{x}{2}\sqrt{c^2-x^2}+\dfrac{c^2}{2}\sin^{-1}\left(\dfrac{x}{c}\right)\right]_0^c\ dy$

$\displaystyle\longrightarrow I=\int\limits_0^a\left(\dfrac{\pi c^2}{4}\right)\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\int\limits_0^ac^2\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\int\limits_0^a\left(a^2-y^2\right)\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\left[a^2y-\dfrac{y^3}{3}\right]_0^a$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\left(a^3-\dfrac{a^3}{3}\right)$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{\pi a^3}{6}}}$

Answer 2

Step-by-step explanation:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

$\sf{ : \implies Area \: of \: rhombus = \dfrac{1}{2} \times d _{1} \times d_{2} }$

$\sf : \implies{ {120 \: cm}^{2} = \dfrac{1}{2} \times 8 \: cm \times x}$

$\sf : \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }$

$\sf : \implies{ {240 \: cm}^{2} = 8 \: cm \times x}$

$\sf : \implies{ \dfrac{240}{8} \: cm = x}$

${ \underline{ \boxed{ \sf \pink{ : \implies{ \bm3 \bm0 \: c m =x}}}}}$

${ \therefore{ \underline{\sf{So \:,the \: length \: of \: other \: diagonal \: is \: 3 0 \: cm}}}}$