# How to evaluate this definite double integral?

$$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$

I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$

But that is incorrect.

• Why do you say that's incorrect? Looks fine to me... – Eli Berkowitz Dec 14 '15 at 3:32
• Your integral looks perfect! Now just evaluate it, try a u-substitution for the inside integral. – Alex Dec 14 '15 at 3:37
• I got -8, I must have made an arithmetic error or something! – d0rmLife Dec 14 '15 at 3:39

Setting $y^2-x^2 = t$, we obtain $-2xdx = dt$. Hence, we obtain $$I = \int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy = \int_0^2 \int_{y^2}^0(-3\sqrt{t}dt)dy = \int_0^2 \int_0^{y^2} 3\sqrt{t}dtdy = \int_0^22y^3 dy = 8$$
• Why does the range of the inner integral change to $y^2,0$ instead of $y^2,y^2-y$ - I thought you put the original range of integration into the u substitution equation? – d0rmLife Dec 14 '15 at 3:47
• @d0rmLife because the first inner integral goes from $x=0$ to $x=y$, and when replacing $y^2-x^2=t$ in the first case you get $y^2-0^2=y^2=t$ and in the second $y^2-y^2=0=t$. When you u-substitute you must change the range accordingly or else it is "wrong": you can get away with it if you evaluate the result in terms of x instead of u in the end, but still it is best to change it. – Fede Poncio Dec 14 '15 at 4:13
Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy$$ $$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy$$ $$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$ $$=\int_{0}^2\left(2y^3\right)\ dy$$ $$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$