จงหาค่าปริพันธ์ต่อไปนี้

โดยการแปลงเป็นระบบพิกัดเชิงขั้ว ได้

• $\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy dx \qquad = \int_0^{2\pi} \int_0^1 r dr d\theta $ วิธีทำ
SageMath

• $\int_{-1}^0\int_{-\sqrt{1-x^2}}^0 \frac{2}{1+\sqrt{x^2+y^2}} dy dx \qquad = \int_{\pi}^{\frac{3\pi}{2}} \int_0^1 \frac{2 r}{1+r} dr d\theta $ วิธีทำ
• $\int_0^2 \int_0^{\sqrt{1-(x-1)^2}} \frac{x+y}{x^2+y^2} dy dx \qquad = \int_0^{\frac{\pi}{2}} \int_0^{2 cos \theta } ( cos \theta +sin \theta ) $   วิธีทำ
• $\int_{-2}^0\int_0^{\sqrt{4-x^2}} \frac{2}{(1+x^2+y^2)^2} dy dx \qquad = \int_{\frac{\pi}{2}}^{\pi} \int_0^2 \frac{2 r}{(1+r^2)^2} dr d\theta   dr d\theta $   วิธีทำ
• $\int_0^1 \int_0 ^{\sqrt{1-y^2}} (x^2+y^2) dx dy \qquad = \int_0^{\frac{\pi}{2}} \int_0^1 r^3 dr d\theta $    วิธีทำ
• $\int_{-1}^1 \int_{-\sqrt{1-y^2}}^0 \frac{4 \sqrt{x^2+y^2}}{1+x^2+y^2} dx dy \qquad = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_0^1 \frac{4 r^2}{1+r^2} dr d\theta $   วิธีทำ
• $\int_0^{ ln 2 } \int_0^{\sqrt{ ln^2 {2} -y^2}} e^{\sqrt{x^2+y^2}} dx dy \qquad = \int_0^{\frac{\pi}{2}} \int_0^{ln 2} r e^r dr d\theta $   วิธีทำ
• $\int_{-1}^1 \int_{-\sqrt{1 -y^2}}^{\sqrt{1 -y^2}} \ln (x^2+y^2+1) dx dy \qquad = \int_0^{2 \pi } \int_0^1 r ln(r^2+1) dr d\theta $   วิธีทำ
• $\int_0^3 \int_0^{\sqrt{9 -y^2}} cos (x^2+y^2) dx dy \qquad = \int_0^{\frac{ \pi}{2}} \int_0^3 r cos(r^2) dr d\theta $  วิธีทำ