How Do You Evaluate a Double Integral with Mixed Powers of x and y?

[cse63146]

Homework Statement

Evaluate the following integral:

\int \int_R x^3 y^4 dx dy

Homework Equations

The Attempt at a Solution

I don't even know where to start. My professor just introduced us to double integrals, now there's a strike going on at my school, so classes are cancelled, and I want to get an early start.

Does anyone know a good link where they do a decent job explaining double integrals? Thank you.

 

[Prologue]

Basically, you consider the 'inside integral' first. Perform the integral int(x^3y^4dx) while considering y^4 as just a constant, then you take the result of that and perform the integral with respect to y.

 

[cse63146]

so in the 'inside integral', the y is treated as a constant?

I forgot to mention that R = [0,2]x[0,5]. What does that mean?

nvm, I got it.

 

[Avodyne]

so in the 'inside integral', the y is treated as a constant?

Yes.

I forgot to mention that R = [0,2]x[0,5]. What does that mean?

It means that x is integrated from 0 to 2, and y is integrated from 0 to 5.

 

[cse63146]

How about this one:

\int\int_R x^3 + y^3 dx dy R = [1,2]x[-1,0]

this is what I got:

\int\int_R x^3 + y^3 dx dy =\int ( \int^2_1 x^3 + y^3 dx) dy = \int (\frac{1}{4}x^4 +xy^3)^{2}_{1} dy = \int^{0}_{-1} \frac{31}{4} + y^3 dy = (\frac{31}{4}y + \frac{1}{4}y^4)^0_{-1} = \frac{15}{2}

but the answer in the back of the book is 7/2, does anyone see what mistake I made? Thank You

 

[xiaochobitz]

er 2^4 is not 32, it's 16~

 

[cse63146]

oh... right. Thanks.

 

[HallsofIvy]

\int\int_R x^3 + y^3 dx dy =\int ( \int^2_1 x^3 + y^3 dx) dy = \int (\frac{1}{4}x^4 +xy^3)^{2}_{1} dy = \int^{0}_{-1} \frac{31}{4} + y^3 dy = (\frac{31}{4}y + \frac{1}{4}y^4)^0_{-1} = \frac{15}{2}
I would recommend that you include all the limits of integration right from the start (so you won't forget!). Also it is a very good idea to include "x= " and "y= " so you won't confuse that:
\int\int_R x^3 + y^3 dx dy =\int_{y= -1}^0 ( \int^2_{x=1} x^3 + y^3 dx) dy = \int_{y=-1}^0 (\frac{1}{4}x^4 +xy^3)^{2}_{1} dy = \int^{0}_{y=-1} \frac{15}{4} + y^3 dy = (\frac{15}{4}y + \frac{1}{4}y^4)^0_{-1} = \frac{7}{2}