Evaluate $\iint(x^{2} + y^{2})dx\,dy$ over the area in the first quadrant bounded by the circle $x^2 + y^2 = a^2$.

Question

Well I am stuck from the beginning. I have two methods:

Method 1:

$y=\sqrt{a^2−x^2}$ here $y$ varies from $\sqrt{a^2−x^2}$ to $0$ $x$ varies from $a$ to $0$ $\iint(x^2+y^2) dx\, dy= \int\left(x^2y+\frac{y^3}{3}\right) dx$ then I do not how to proceed

Method 2: Changing into polar coordinates. $$\text{Let}\ x=r\cosθ, y=r\sinθ,$$ $$x^2+y^2= r^2$$ hence $r$ varies from $a$ to $0$; $θ$ varies from $0$ to $90$

\begin{align*} J(x,\frac{y}{r},θ)&= r\\ \iint(x^2+y^2) dx\, dy&= \iint(x^2+y^2)J dr\, dθ\\ &= \iint (r^2\cdot r)\ dr\, dθ\\ &= \iint r^3 dr\, dθ\\ &= \int \frac{r^4}{4} dθ\\ &= \int \frac{a^4}{4} dθ\\ &= \frac{a^4}{4}\cdot\frac{\pi}{2} \end{align*}

But that is not the answer. The answer is actually said to be $\frac{a^5}{5}$. Could anyone please help me?

Answer 1

If you make the change of variable $x = r\cos(\theta)$ and $y = r\sin(\theta)$, the proposed integral reduces to \begin{align*} \iint_{R}(x^{2} + y^{2})\mathrm{d}x\mathrm{d}y = \int_{0}^{\pi/2}\int_{0}^{a}r^{3}\mathrm{d}r\mathrm{d}\theta = \frac{\pi a^{4}}{8} \end{align*}

So your calculations are correct!

Answer 2

About first method. You need to add the limits of integration

$$\iint_{x^2+y^2<a^2 \\x>0,y>0}(x^2+y^2) dx\, dy=\int_{0}^{a} \int_{0}^{\sqrt{a^2-x^2}} (x^2+y^2)dy\,dx=\\ \int_0^a \left.\left(x^2y+\frac{y^3}{3}\right)\right|_{y=0}^{y=\sqrt{a^2-x^2}}dx=\int_0^a \left(x^2\sqrt{a^2-x^2}+\frac{1}{3}(a^2-x^2)\sqrt{a^2-x^2}\right)dx=\\ \int_0^a \left(\frac{2}{3}x^2\sqrt{a^2-x^2}+\frac{1}{3}a^2\sqrt{a^2-x^2}\right)dx$$

Standard substitution for such integrals is $x=a\sin t$ and this leads to the same result as in your second method but needs more calculation:

$$\int_0^a \left(\frac{2}{3}x^2\sqrt{a^2-x^2}+\frac{1}{3}a^2\sqrt{a^2-x^2}\right)dx=\\ \int_0^{\pi/2}\left(\frac{2}{3}a^3\sin^2 t \cos t+\frac{1}{3}a^3\cos t\right)a\cos t dt=\\ a^4\int_0^{\pi/2}\left(\frac{2}{3}\sin^2 t \cos^2 t+\frac{1} {3}\cos^2 t\right) dt=\\ a^4\int_0^{\pi/2}\left(\frac{1}{6}\sin^2 2t+\frac{1}{6}\cos 2t+\frac{1}{6}\right) dt=\\ a^4\int_0^{\pi/2}\left(\frac{1}{12}-\frac{1}{12}\cos 4t+\frac{1}{6}\cos 2t+\frac{1}{6}\right) dt=\\ a^4\int_0^{\pi/2}\left(\frac{1}{4}-\frac{1}{12}\cos 4t+\frac{1}{6}\cos 2t\right) dt=\\ a^4\left.\left(\frac{t}{4}-\frac{1}{48}\sin 4t+\frac{1}{12}\sin 2t\right)\right|_{t=0}^{t=\pi/2}=\frac{\pi a^4}{8}$$

Answer 3

Method 1:

$y=\sqrt{a^2−x^2}$ here $y$ varies from $\sqrt{a^2−x^2}$ to $0$ $x$ varies from $a$ to $0$ $\iint(x^2+y^2) dx\, dy= \int\left(x^2y+\frac{y^3}{3}\right) dx$ then I do not how to proceed

You do the innermost integral first, and you put the limits of integration into it to remove one of the variables. Let $b\equiv \sqrt{a^2-y^2}$ then

$$ \int_{0}^a\int_{0}^b\left(x^2+y^2\right) dx \;dy = \int_{0}^a\left[{x^3\over 3}+x y^2\right]_{0}^b dy\\ =\int_{0}^a b\left({b^2\over 3}+y^2\right) dy $$

Substitute $y=a\sin t$, then $dy=a\cos t\;dt$ and $b=a\cos t$ and you get $$ {1\over 3}a^4\int \cos^2 t(\cos^2 t+3\sin^2t) dt $$ Now $$ \int \cos^2 t(\cos^2 t+3\sin^2t) dt=\int \cos^2t+2(\cos t\sin t)^2 dt\\ =\frac12t-\frac14\sin2t +\frac14t-\frac1{16}\sin 4t +C $$ Here I have used the double-angle formulas repeatedly.

Substitute in the limits and you get the same answer.