# Double Integral Practice Problems with Answers and Solutions

> 12 double integral practice problems with answers: iterated integrals, general regions, switching the order, polar coordinates, volume and average value.

Source: https://practicecalculus.com/practice-problems/double-integrals
Updated: 2026-11-06

$$ \iint_R f(x, y)\,dA = \int_a^b\!\int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 3
- **You need:** [Definite integrals](/practice-problems/definite-integrals) and [u-substitution](/practice-problems/u-substitution)
- **Main idea:** Integrate the inside variable first, treating the outer variable as a constant, then integrate again
- **Practice with new numbers:** [Iterated integrals lesson](/learn/calculus-3/multiple-integrals/iterated-integrals)

## Key takeaways

- Work from the inside out. The inner limits can depend on the outer variable, but the outer limits must be numbers.
- If the inner integral has no elementary antiderivative, sketch the region and switch the order.
- For disks and circles, switch to polar coordinates and remember $dA = r\,dr\,d\theta$.

These **double integral practice problems** start with rectangles and move on to general regions, switching the order of integration, polar coordinates, volume and average value. Try each one first. The answers come first, then full solutions.

The setup for a region between two curves:

$$\iint_R f(x, y)\,dA = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\right)dx$$

Do the inside integral first, treating $x$ as a constant. For more on regions and order, read the [double integrals guide](/blog/double-integrals).

## Problems

**Rectangles**

1. $\displaystyle\int_0^2\!\int_0^1(x + y)\,dy\,dx$
2. $\displaystyle\iint_R 6xy^2\,dA$, $R = [0, 2]\times[0, 1]$
3. $\displaystyle\int_0^{\pi}\!\int_0^1 y\sin x\,dy\,dx$

**General regions**

4. $\displaystyle\int_0^1\!\int_0^x xy\,dy\,dx$
5. $\displaystyle\iint_R x\,dA$, where $R$ is between $y = x^2$ and $y = x$
6. Find the area between $y = x^2$ and $y = 4$ with a double integral.

**Switch the order**

7. $\displaystyle\int_0^1\!\int_y^1 e^{x^2}\,dx\,dy$
8. $\displaystyle\int_0^1\!\int_x^1\sin(y^2)\,dy\,dx$

**Polar coordinates**

9. $\displaystyle\iint_D(x^2 + y^2)\,dA$, $D$ the unit disk
10. $\displaystyle\iint_D e^{-(x^2 + y^2)}\,dA$, $D$ the disk $x^2 + y^2 \le 4$
11. Find the volume under $z = 4 - x^2 - y^2$ and above the $xy$-plane.

**Average value**

12. Find the average value of $f(x, y) = xy$ on the square $[0, 2]\times[0, 2]$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $3$ |
| 2 | $4$ |
| 3 | $1$ |
| 4 | $\frac{1}{8}$ |
| 5 | $\frac{1}{12}$ |
| 6 | $\frac{32}{3}$ |
| 7 | $\frac{e - 1}{2} \approx 0.859$ |
| 8 | $\frac{1 - \cos 1}{2} \approx 0.230$ |
| 9 | $\frac{\pi}{2}$ |
| 10 | $\pi\left(1 - e^{-4}\right) \approx 3.084$ |
| 11 | $8\pi$ |
| 12 | $1$ |

## Step-by-step solutions

**1.** Inside: $\int_0^1(x + y)\,dy = x + \frac{1}{2}$. Outside:

$$\int_0^2\left(x + \frac{1}{2}\right)dx = 2 + 1 = 3$$

**2.** On a rectangle, a product $g(x)h(y)$ splits into two single integrals:

$$\int_0^2 6x\,dx\cdot\int_0^1 y^2\,dy = 12\cdot\frac{1}{3} = 4$$

**3.** Split again: $\int_0^{\pi}\sin x\,dx\cdot\int_0^1 y\,dy = 2\cdot\frac{1}{2} = 1$.

**4.** Inside: $\int_0^x xy\,dy = x\cdot\frac{x^2}{2} = \frac{x^3}{2}$. Outside: $\int_0^1\frac{x^3}{2}\,dx = \frac{1}{8}$.

**5.** The curves meet at $x = 0$ and $x = 1$, with $y = x$ on top:

$$\int_0^1\!\int_{x^2}^{x}x\,dy\,dx = \int_0^1 x(x - x^2)\,dx = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$$

**6.** With $f = 1$, the double integral is the area:

$$\int_{-2}^{2}\!\int_{x^2}^{4}1\,dy\,dx = \int_{-2}^{2}(4 - x^2)\,dx = \frac{32}{3}$$

**7.** $e^{x^2}$ has no elementary antiderivative, so switch. The region is $0 \le y \le x \le 1$, a triangle. In the other order, $x$ runs from $0$ to $1$ and $y$ from $0$ to $x$:

$$\int_0^1\!\int_0^x e^{x^2}\,dy\,dx = \int_0^1 x\,e^{x^2}\,dx = \frac{e - 1}{2}$$

**8.** The region is $0 \le x \le y \le 1$. Switch so $y$ is outside:

$$\int_0^1\!\int_0^y\sin(y^2)\,dx\,dy = \int_0^1 y\sin(y^2)\,dy = \left[-\frac{\cos(y^2)}{2}\right]_0^1 = \frac{1 - \cos 1}{2}$$

**9.** In polar, $x^2 + y^2 = r^2$ and $dA = r\,dr\,d\theta$:

$$\int_0^{2\pi}\!\int_0^1 r^2\cdot r\,dr\,d\theta = 2\pi\cdot\frac{1}{4} = \frac{\pi}{2}$$

**10.** With $u = r^2$:

$$\int_0^{2\pi}\!\int_0^2 e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\frac{1}{2}\left(1 - e^{-4}\right) = \pi\left(1 - e^{-4}\right)$$

**11.** The surface meets the plane where $r = 2$. In polar:

$$\int_0^{2\pi}\!\int_0^2(4 - r^2)\,r\,dr\,d\theta = 2\pi\left[2r^2 - \frac{r^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi$$

**12.** $\iint xy\,dA = \int_0^2 x\,dx\cdot\int_0^2 y\,dy = 2\cdot 2 = 4$. The area of the square is $4$, so the average is $\frac{4}{4} = 1$.

## Common mistakes on these problems

- **Outer limits with variables.** After both integrals the answer must be a number. If an $x$ is left over, the limits are in the wrong order.
- **Switching order without redrawing.** In problem 7, swapping the integral signs and keeping the limits is wrong. Sketch the region and read new limits from it.
- **Forgetting $r$ in polar.** Problem 9 without the extra $r$ gives $\frac{2\pi}{3}$ instead of $\frac{\pi}{2}$.
- **Treating the outer variable as changing.** In the inner integral of problem 4, $x$ is a constant, so $\int_0^x xy\,dy = \frac{x^3}{2}$.

## Keep practicing

For unlimited problems with instant feedback, try the [iterated integrals lesson](/learn/calculus-3/multiple-integrals/iterated-integrals), the [switching the order lesson](/learn/calculus-3/multiple-integrals/switching-the-order), the [general regions lesson](/learn/calculus-3/multiple-integrals/general-regions) and the [polar double integrals lesson](/learn/calculus-3/multiple-integrals/polar-double-integrals).

## Frequently asked questions

### How do you know which order to integrate in?

For a rectangle with nice functions, either order works. For other regions, pick the order that gives simpler limits, or the one where the inner integral can actually be done. Problems 7 and 8 can only be done after switching.

### Why is there an extra r in polar double integrals?

A small polar region is nearly a rectangle with sides $dr$ and $r\,d\theta$, so its area is $r\,dr\,d\theta$. Leaving out the $r$ is the most common polar mistake.

### What does a double integral represent?

If $f \ge 0$, it is the volume under the surface $z = f(x, y)$ above the region. With $f = 1$, it is the area of the region. Divided by the area, it is the average value of $f$.