Review Problems for Test III - Part I

The University of Akron
Department of Mathematics
Calculus III
Spring 2026, Golovaty

Review Problems for Test III - Part I

8. Compute the following double integrals.

$\displaystyle\mathop{\int\!\!\!\int}_{R}{\left(x\sin{y}-y\sin{x}\right)}\,dA$ , where $R:=\left\{(x\,,y):0\leq x\leq\frac{\pi}{2}\,,\ 0\leq x\leq\frac{\pi}{3}\right\}\,.$
$\displaystyle\mathop{\int\!\!\!\int}_D{x^2}\,dA$ , where is the region bounded by $y=\frac{16}{x}$ , , and .
$\displaystyle\mathop{\int\!\!\!\int}_D{x{\left(1+y^2\right)}^{-1/2}}\,dA$ , where is the region in the first quadrant enclosed by , , and .
$\displaystyle\mathop{\int\!\!\!\int}_R{\left(x-1\right)}\,dA$ , where is the region in the first quadrant enclosed between and .
$\displaystyle{\mathop{\int\!\!\!\int}_R\sin{\left(y^3\right)}}\,dA$ , where is the region bounded by $y=\sqrt{x}$ , , and .

9. Use double integration to find the area of the plane region enclosed between the curves $y=\sin{x}$ and $y=\cos{x}$ , for $0\leq x\leq\frac{\pi}{4}$ .

10. Use double integration to find the volume of each solid.

The solid in the first octant bounded above by the paraboloid , below the plane , and laterally by and .
The wedge cut from the cylinder by the planes and .

11. Evaluate the integral by first reversing the order of integration.

$\displaystyle\int_0^4\!\int_{\sqrt{y}}^2e^{x^3}\,dx\,dy\,.$
$\displaystyle\int_1^3\!\int_0^{\ln{x}}x\,dy\,dx\,.$

12. Use the double integral in polar coordinates to find the area of a given region.

The region enclosed by the cardioid $r=1-\cos{\theta}$ .
The region inside the circle $r=4\sin{\theta}$ and outside of the circle .

13. Use the polar coordinates to evaluate the double integral.

$\displaystyle{\mathop{\int\!\!\!\int}_R{e^{-x^2-y^2}}}\,dA$ , where is the region inside by the circle .
$\displaystyle{\mathop{\int\!\!\!\int}_R{2y}}\,dA$ , where is the region in the first quadrant bounded below by the line and above by the circle ${(x-1)}^2+y^2=1$ .

14. Evaluate the integral by reverting to polar coordinates

$\displaystyle\int_{0}^{1}\!\int_{0}^{\sqrt{1-y^2}}{\cos{(x^2+y^2)}}\,dx\,dy$
$\displaystyle\int_{0}^{1}\!\int_{y}^{\sqrt{y}}{\sqrt{x^2+y^2}}\,dx\,dy$

Dmitry Golovaty