Double integral of x + 2y over the region between y = x^2 and y = 2… Step by Step

Step by step

The question

Question: find the double integral of x + 2y over the region between y = x² and y = 2 − x².

∫ ∫ (x + 2y) dA, between y = x² and y = 2 − x²

1 Draw the region

Draw the grid.

Draw y = 2 − x²

and y = x².

The region is between them.

Where do they meet?

Where the two heights are equal.

2 − x² = x²

Bring everything to one side.

2 − 2x² = 0

Move the number and divide: x² = 1.

x² = 1

So they meet at x = −1 and x = 1.

x = −1 or x = 1

Which is on top?

Try x = 0: y = 2 − x² is higher.

x = 0: 2 − x² → 2, x² → 0

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 2 − x².

Outside, x goes from −1 to 1.

∫ ∫ (x + 2y) dA = ∫ from −1 to 1 of ∫ from x² to 2 − x² of (x + 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 2 − x² of (x + 2y) dy = [xy + y²] from x² to 2 − x²

Put in the top y = 2 − x², then take away the bottom y = x².

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

Work out each term on its own: multiply out the brackets.

∫ from x² to 2 − x² of (x + 2y) dy = ((2x − x³) + (4 − 4x² + x⁴)) − (x³ + x⁴)

Add the like terms in each part.

∫ from x² to 2 − x² of (x + 2y) dy = (2x − x³ + 4 − 4x² + x⁴) − (x³ + x⁴)

So it is 2x − 2x³ + 4 − 4x².

∫ from x² to 2 − x² of (x + 2y) dy = 2x − 2x³ + 4 − 4x²

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 −1 to 1 of (2x − 2x³ + 4 − 4x²) dx = [x² − x⁴2 + 4x − 4x³3] from −1 to 1

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

∫ from −1 to 1 of (2x − 2x³ + 4 − 4x²) dx = ((1)² − (1)⁴2 + 4(1) − 4(1)³3) − ((−1)² − (−1)⁴2 + 4(−1) − 4(−1)³3)

So it is 16/3.

∫ from −1 to 1 of (2x − 2x³ + 4 − 4x²) dx = 163

5 Check

Let's check if it makes sense.

The region's area is 8/3.

On it, f goes from −0.13 to 4.13.

area = 8/3, f from −0.13 to 4.13

So the answer must be between −0.333 and 11.

5.333 is.

Right!

−0.333 ≤ 5.333 ≤ 11

Let's look back.

First: draw the region.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer