# Recitation 10

Section 12.1 Problem 24: Calculate the double integral. $$\iint_R\frac{1+x^2}{1+y^2}dA, R=\{(x,y)|0\leq x\leq1, 0\leq y\leq 1\}$$

Solution: We can decompose the double integral into the product of two integrals $$\int_0^1(1+x^2)dx\int_0^1\frac{1}{1+y^2}dy=(1+\frac{1}{3})\frac{\pi}{4}=\frac{\pi}{3}.$$

Section 12.1 Problem 35: Find the volume of the solid enclosed by the paraboloid $z=2+x^2 +(y-2)^2$ and the planes $z=1,x=1, x=-1,y=0$,and $y=4$.

Comment: The volume is given by the formula $$\int_{[-1,1]\times[0,4]}\left(2+x^2+(y-2)^2-1\right)dA.$$

Section 12.2 Problem 15: Evaluate the double integral. $$\iint_D x\cos y dA, D\text{ is bounded by } y=0, y=x^2, x=1$$

Comment: The double integral is equal to $$\int_0^1x\cdot\left(\int_0^{x^2}\cos y\right)dx.$$

Section 12.2 Problem 29: Find the volume of the given solid bounded by the cylinder $x^2 + y^2= 1$ and the planes $y = z, x = 0, z = 0$ in the first octant.

Solution: The volume is given by the double integral $$\int_D y dA,$$ where the region $D$ is bounded by the circle $x^2+y^2=1$ in the first quadrant of the $xy$ plane. We can rewrite this double integral by $$\int_0^1\left(\int_0^{\sqrt{1-x^2}}ydy\right)dx=\int_0^1\frac{1-x^2}{2}dx=\frac{1}{3}.$$