Question

Evaluate:

a) ∫∫D (e^(y^2) + 1) dA where D is the triangle with vertices (0,0), (-2,4) and (8,4).

b) ∫∫D x^(5)sin(y^4) dA where D is the region in the 2nd quadrant bounded by y =3x^2, y = 12 and the y-axis.

Answer 1

a $\int_0^4 \int_{-2}^8(e^{y^2}+1)dxdy= \int_0^4(xe^{y^2}+x)|_{-2}^8= \int_0^4(10e^{y^2}+10+C)dy=5\sqrt\pi erfi(y)+y(10+C)|_0^4=5\sqrt{\pi} erfi(4)+4(10+C)+ C_1$

b.

$\int_{3x^2}^{12}\int_{(-12)}^0 x^5siny^4 dxdy=\int_{3x^2}^{12}siny^4x^6/6+C|_{-12}^0 dy=497664C\int_{3x^2}^{12}siny^4 dy=497664C_1 \dfrac{\left(\operatorname{\Gamma}\left(\frac{1}{4},\mathrm{i}x^8\right)+\operatorname{\Gamma}\left(\frac{1}{4},-\mathrm{i}x^8\right)-\operatorname{\Gamma}\left(\frac{1}{4},20736\mathrm{i}\right)-\operatorname{\Gamma}\left(\frac{1}{4},-20736\mathrm{i}\right)\right)\sin\left(\frac{{\pi}}{8}\right)+\left(\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{4},\mathrm{i}x^8\right)-\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{4},-\mathrm{i}x^8\right)-\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{4},20736\mathrm{i}\right)+\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{4},-20736\mathrm{i}\right)\right)\cos\left(\frac{{\pi}}{8}\right)}{8}$