Question

Please solve this double integration.....​

Answer 1

Topic:

Integration

Solution:

We need to evaluate the given integral.

$\displaystyle\sf I = \int_0^\pi\int_{2\sin\theta}^{4\sin\theta} r^3\, dr\, d\theta$

Solving the inner integral w.r.t. r, we get:
$\displaystyle\sf I = \int_0^\pi \bigg[\dfrac14r^4\bigg]^{4\sin\theta}_{2\sin\theta}\, d\theta$

$\displaystyle\sf I =\dfrac14 \int_0^\pi \bigg[r^4\bigg]^{4\sin\theta}_{2\sin\theta}\, d\theta$

$\displaystyle\sf I = \dfrac14\int_0^\pi \bigg[(4\sin\theta)^4 - (2\sin\theta)^4\bigg]\, d\theta$

$\displaystyle\sf I = \dfrac14\int_0^\pi \bigg[(4^4\sin^4\theta) - (2^4\sin^4\theta)\bigg]\, d\theta$

$\displaystyle\sf I = \dfrac14\int_0^\pi \bigg[\sin^4\theta(4^4 - 2^4) \bigg]\, d\theta$

$\displaystyle\sf I =\int_0^\pi 60\sin^4(\theta) d\theta$

$\displaystyle\sf I =\int_0^\pi 60\left(\dfrac{1 - \cos2\theta}{2}\right)^2 d\theta\qquad\left\{\because2\sin^2(x) = 1-\cos(2x)\right\}$

$\displaystyle\sf I =15\int_0^\pi( 1 - \cos2\theta)^2\, d\theta$

$\displaystyle\sf I =15\int_0^\pi1 + \cos^22\theta - 2\cos2\theta\, d\theta$

${\displaystyle\sf I =15\int_0^\pi1 + \dfrac{1 + \cos4\theta}{2} - 2\cos2\theta\, d\theta\qquad\left\{\because 2\cos^2(x) = 1 + cos(2x)\right\}}$

${\displaystyle\sf I =15\int_0^\pi\ d\theta +\dfrac{15}{2}\int_0^\pi1 + \cos4\theta\ d\theta-30\int_0^\pi \cos2\theta\, d\theta}$

${\displaystyle\sf I =15\pi +\dfrac{15}{2}\bigg[\theta + \dfrac{\sin4\theta}{4}\bigg]^{\pi}_0-30\bigg[\dfrac{\sin2\theta}{2}\bigg]^\pi_0}$

${\displaystyle\sf I =15\pi +\dfrac{15\pi}{2}}$

${\displaystyle\sf I =\dfrac{45\pi}{2}}$

So the required answer is,

$\underline{\boxed{ \int_0^\pi\int_{2\sin\theta}^{4\sin\theta} r^3\, dr\, d\theta = \dfrac{45\pi}{2}}}$