Exercises on Double integrals

Evaluate the following double integrals, using if necessary appropriate symmetry considerations.

1. ∫∫_{[0,1] x [0, π/2]} y sin(xy) dx dy

2. ∫∫_T (x + sin y) dx dy

   with T = {(x,y): 0 ≤ x < 1; 0 ≤ y ≤ 1 − x}.

3. ∫∫_{x² + y² < 2} xy³ dx dy

Solutions

1. The Integrand is continuous on the rectangular domain, and Fubini's theorem can be applied. We use the substituion u = xy.

   $${\int }_0^{\pi /2}({\int }_0^{1}y\sin (xy)\text{d}x)\text{d}y={\int }_0^{\pi /2}({\int }_0^y\sin u\text{d}u)\text{d}y={\int }_0^{\pi /2}{[-\cos u]}_0^y\text{d}y={\int }_0^{\pi /2}(-\cos y+1)\text{d}y={[-\sin y+y]}_0^{\pi /2}=-1+\frac{\pi }{2}=0.570796$$

   We could have also computed the y-integral first and get the same result. But the we would have faces the integrand usinu wich requires integration by parts to compute by hand. In general is also possible that the integration followin a particular order canno be computed by hand, while picking the right order yields an integral that is relatevely easy to compute.  ■

2. $${\int }_0^1({\int }_0^{1-x}(x+\sin y)\text{d}x)\text{d}y={\int }_0^1{[xy+-\cos y]}_0^{1-x}\text{d}x={\int }_0^1[x(1-x)-\cos (1-x)+1]\text{d}x={[\frac{x}{2}-\frac{x^3}{3}+\sin (1-x)+x]}_0^1=\frac{7}{6}-\sin 1\approx 0.325196$$

3. The domain can be written as D = {(x,y): −√2 ≤ x < √2; −√(2 + x²) ≤ y ≤ √(2 + x²)}.

$${\int }_{-\sqrt{2}}^{\sqrt{2}}\left({\int }_{-\sqrt{2+x^2}}^{\sqrt{2+x^2}}x{y}^3\text{d}y\right)\text{d}x={\int }_{-\sqrt{2}}^{\sqrt{2}}{\left[x\frac{y^4}{4}\right]}_{-\sqrt{2+x^2}}^{\sqrt{2+x^2}}\text{d}x=0$$

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