0
$\begingroup$

$\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

Okay so I converted this into polar form because I was told to do so I got the integral of $(7r\cos\theta + 3r\sin\theta)rdrd\theta$ where $0\le \theta \le \pi/4$ and $0\le r \le \sqrt{18}$

I think I'm making a mistake solving this integral. I keep getting $72$ which is incorrect.

Work: Taking the first integral I get $7r^3\cos\theta/3 + r^3\sin\theta$ from $0$ to $\sqrt{18}$ then I get $7(\sqrt{18})^3 \cos\theta/ 3 + \sqrt {18}^3\sin\theta$ Then I took the second integral and got $7(\sqrt{18})^3 \sin\theta/ 3 - (\sqrt {18})^3\cos{\theta}$ I plugged in the values of pi/4 and o and got

$126 - 54 + (\sqrt{18}^3) = 148.367532368147$

Sorry, while typing the work I found my mistake. Thanks everyone for trying to help me out :)

$\endgroup$
  • $\begingroup$ Can you show all your working? $\endgroup$ – Calvin Lin Feb 24 '14 at 6:25
  • $\begingroup$ Sure, I'll edit it $\endgroup$ – ayv2 Feb 24 '14 at 6:26
  • 1
    $\begingroup$ Glad you found your mistake. By the way, the result is simply $72+54 \sqrt{2}$ the value of which coinciding with your number. $\endgroup$ – Claude Leibovici Feb 24 '14 at 6:36
0
$\begingroup$

Let $I$ denote the integral value. By calculating the integral by the polar form, we have: \begin{equation} I=\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)+3r^2\sin(\theta)drd\theta\\ =\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)drd\theta+\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}3r^2\sin(\theta)drd\theta\\ =\int_0^{\frac{\pi}{4}}\cos(\theta)d\theta\int_0^\sqrt{18}7r^2dr+\int_0^{\frac{\pi}{4}}\sin(\theta)d\theta\int_0^\sqrt{18}3r^2dr\\ =\frac{\sqrt{2}}{2}\times126\sqrt{2}+(1-\frac{\sqrt{2}}{2})\times54\sqrt{2}\\ =72+54\sqrt{2} \end{equation}

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

I'm not sure about polar form, but you can evaluate the integral as such:

$\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

$\int\limits_{0}^{3}(\frac{7}{2}x^2 + 3y\cdot x)|_{\sqrt{18-y^2}}^y$ $dy$

$\int\limits_{0}^{3}(\frac{7}{2}y^2 + 3y^2))-(\frac{7}{2}(18-y^2) + 3y\cdot \sqrt{18-y^2})$ $dy$

$\int\limits_{0}^{3} 10\cdot y^2-3y\cdot \sqrt{18-y^2}-63$ $dy$

It should be relatively easy to solve from there using more basic, one-variable methods.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have to do the problem in polar form... $\endgroup$ – ayv2 Feb 24 '14 at 6:25
  • $\begingroup$ Ah. Good luck, then. $\endgroup$ – walkar Feb 24 '14 at 6:34
  • $\begingroup$ Dear @walkar, maybe you mistake the up and low boundary of integral in the 2nd line of your deduction. $\endgroup$ – Lion Feb 24 '14 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.