Question

$\textsf{Evaluate the integral:}\\\\\displaystyle \int\limits^3_0 \int\limits^{\sqrt{9-x^2}}_0 (x^2+y^2)^{\dfrac{3}{2}} dydx$

Answer 1

it can be solved by using polar co - ordinates.

Let $x=rsin\theta$ and $y=rcos\theta$

so, x² + y² = $r^2(cos^2\theta+sin^2\theta)=r^2$

now, differentiate both sides,

2xdx + 2ydy = 2rdr

or, xdx + ydy = rdr......(1)

$\frac{y}{x}=tan\theta$

differentiate both sides,

$\frac{xdy-ydx}{x^2}=sec^2\theta.d\theta$

$xdy-ydx=x^2sec^2\theta.d\theta$.....(2)

multiply equations (1) and (2),

you get, (x²+y^²)dy. dx = $x^2sec^2\theta.d\theta.rdr$

$r^2dy.dx = r^2cos^2\theta. sec^2\theta. rdr. d\theta$

so, $dy.dy=rdr.d\theta$ now in next time you can use this directly.

draw graph of y = √(9 - x²) or y² + x² = 3² it is a circle right?,

see shaded region , this is our region in which we have to integrate the expression.

here, r varies from 0 to 3 [ I mean minimum value of r = 0, and max = 3 ]

$\theta$ varies from 0 to π/2 [ see figure to understand, ]

now integration converted into $\int\limits^{\pi/2}_0\int\limits^3_0{r^3}rdr. d\theta$

or, $\int\limits^{\pi/2}_0{\left[\frac{r^5}{5}\right]^3_0}\,d\theta$

or, $\frac{243}{5}\int\limits^{\pi/2}_0{d\theta}$

or, $\frac{243\pi}{10}$ it is the required answer .

Answer 2

Answer:

what we find please mark brainliest