Steply

Double integral of 2y over 0 ≤ x ≤ 1, x^2 ≤ y ≤ 1

Step by step, with the reason for every step · Double integrals over general regions · Calculus 3

Step by step

The question

Question: find the double integral of 2y/0 ≤ x ≤ 1, x² ≤ y ≤ 1.

∫ ∫ 2y dA, 0 ≤ x ≤ 1, x2 ≤ y ≤ 1

1 Draw the region

Draw the grid.

Draw y = 1

and y = x².

The region is between them.

2 Find the limits

Take a thin upright strip.

On it, y goes from the bottom curve up to the top curve.

So inside, y goes from x² to 1.

Outside, x goes from 0 to 1.

∫ ∫ 2y dA = ∫ from 0 to 1 of ∫ from x² to 1 of 2y dy dx

Careful!

The inside integral is dy, so its limits are the y limits.

They may have x in them.

The outside limits are only numbers.

Careful: dy inside: the y limits go inside

3 The inner integral

Inside first: integrate in y.

x stays fixed, like a plain number.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from x² to 1 of 2y dy = y2 from x² to 1

Put in the top y = 1, then take away the bottom y = x².

∫ from x² to 1 of 2y dy = (1)2 − (x2)2

So it is 1 − x⁴.

∫ from x² to 1 of 2y dy = 1 − x4

4 The outer integral

Now the outside integral, in x.

Its limits are numbers.

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 1 of (1 − x4) dx = [x − x55] from 0 to 1

Put in the top x = 1, then take away the bottom x = 0.

∫ from 0 to 1 of (1 − x4) dx = ((1) − (1)55) − ((0) − (0)55)

The bottom part is 0.

So it is 4/5.

∫ from 0 to 1 of (1 − x4) dx = 45

5 Check

Let's check if it makes sense.

The region's area is 2/3.

On it, f goes from 0 to 2.

area = 2/3, f from 0 to 2

So the answer must be between 0 and 1.333.

0.8 is.

Right!

0 ≤ 0.8 ≤ 1.333

Let's look back.

First: draw the region.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer
∫ from 0 to 1 of (1 − x4) dx = 45

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ y dA, 0 ≤ x ≤ 1, x2 ≤ y ≤ 1
Show the answer
25

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