[12.2] - Double Integrals - Practice
- $\int_0^2\int_0^2(x^2-y^2)\,dy\,dx$
- $\int_0^{\frac{\pi}{4}}\int_{y=0}^{\frac{\pi}{2}}\cos(2x+y)\,dy\,dx$
Hint: According to one of the Trig addition formulas, $\cos(\alpha+\beta)= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$
- $\int_0^2\int_{y=1}^3x^3y\,dy\,dx$
- $\int_{-1}^{1}\int_{y=0}^{\pi}x^2\sin y\,dy\,dx$
- $\int_0^2\int_{y=x^2}^{2x}(x^2+2y)\,dy\,dx$
- $\int_0^3\int_{y=0}^{9-x^2}4x\,dy\,dx$
Answers: 1.) 0 2.) 0 3.) 16 4.) $\frac 43$ 5.) $\frac{88}{15}$ 6.) 81 if (! $homepage){ $stylesheet="/~paulmr/class/comments.css"; if (file_exists("/home/httpd/html/cment/comments.h")){ include "/home/httpd/html/cment/comments.h"; } } ?>