Double Integral Practice Problems with Answers and Solutions

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:

Do the inside integral first, treating x as a constant. For more on regions and order, read the double integrals guide.

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

Step-by-step solutions

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

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

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:

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

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:

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

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

10. With u = r^2:

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

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, the switching the order lesson, the general regions lesson and the polar double integrals lesson.