Double Integral 4x^3 dA: Evaluating Region Bounded by y=(x-1)^2 and y=-x+3

[Math10]

Homework Statement

Evaluate the double integral 4x^3 dA, where R is the region bounded by the graphs of y=(x-1)^2 and y= -x+3.

Homework Equations

None.

The Attempt at a Solution

The answer in the book is 72/5. But I got 45 as the answer.
Here's my work:
(x-1)^2= -x+3
x^2-2x+1= -x+3
x=2, -1
y=1, 4.
double integral of 4x^3 dx dy from -1 to 2 for dx and from 1 to 4 for dy
integral of x^4 (evaluate from -1 to 2) dy from 1 to 4 for dy
=45
Is my answer correct or the book's answer is right?

 

[Mark44]

Homework Statement

Evaluate the double integral 4x^3 dA, where R is the region bounded by the graphs of y=(x-1)^2 and y= -x+3.

Homework Equations

None.

The Attempt at a Solution

The answer in the book is 72/5. But I got 45 as the answer.
Here's my work:
(x-1)^2= -x+3
x^2-2x+1= -x+3
x=2, -1
y=1, 4.
double integral of 4x^3 dx dy from -1 to 2 for dx and from 1 to 4 for dy

No. The y values run from the parabola up to the line. They do not run from 1 to 4.

integral of x^4 (evaluate from -1 to 2) dy from 1 to 4 for dy
=45
Is my answer correct or the book's answer is right?

For sure your answer is incorrect.

 

[Math10]

So how do I find the limits of integration for y?

 

[Mark44]

R is the region bounded by the graphs of y=(x-1)^2 and y= -x+3.

You get the limits from these equations. Did you draw a picture of this region?

Also, I got 72/5 as my answer, same as your book.

 

[Math10]

Since it's a double integral, the x is from -1 to 2 for dx. What about y?

 

[O_o]

You can see easily what the region is by plotting your boundary functions here:

http://www.mathsisfun.com/data/function-grapher.php

When x is between -1 and 2, what is the upper and lower bound of y? You can get it just by looking at the graph

 

[Math10]

I still don't get it. What are the limits of integration for y?

 

[O_o]

When x is between -1 and 2 one of the functions is lower than the other. When you plot them can you see which it is? That is the lower bound. And the upper bound is the function that is higher.

The upper and lower bound varies with x. That's why you can't just use constants...you need to write the bounds in terms of functions. Try watching these 2 videos:

https://www.khanacademy.org/math/mu...tegrals/double_integrals/v/double-integrals-5

https://www.khanacademy.org/math/mu...tegrals/double_integrals/v/double-integrals-6

 

[Mark44]

I still don't get it. What are the limits of integration for y?

I asked before, but you didn't respond. Did you sketch a graph of the region over which you're integrating?

 

[Math10]

You know what? I got it.