**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}.