MHB How do we evaluate the double integral with $15x^2$ inside and all the R values?

[karush]

ok just seeing if I have this set up correctly before evaluate..

View attachment 7256
where does $15x^2$ come from?
if $15x^2$ is inside this why would we need all the R values

 

[MarkFL]

Re: 15.2.51 double integral

You have the order of integration correct, at least if you want to state the volume as one iterated integral. Your limits of integration aren't correct though. You are using horizontal strips, and so you want the lower limit to lie on the line:

$$y=2x+4$$

But, you want the $x$-value here, so you want to solve for $x$, to obtain:

$$x=\frac{y-4}{2}$$

The upper limit lies on the curve:

$$y=x^3\implies x=y^{\frac{1}{3}}$$

And thus, we have:

$$\frac{y-4}{2}\le x\le y^{\frac{1}{3}}$$

Your limits for the outer integral are correct, thus we have:

$$I=\int_0^8\int_{\frac{y-4}{2}}^{y^{\frac{1}{3}}} 15x^2\,dx\,dy$$

The integrand $15x^2$ was given, and it represents a $z$-value. Since the integrand is non-negative over $R$, you can think of this iterated integral as representing the volume of a 3-dimensional object. The integrand here tells you how high the object extends above the $xy$-plane.

 

[karush]

Re: 15.2.51 double integral

ok next step !
$\displaystyle I=\int_0^8\int_{\frac{y-4}{2}}^{y^{\frac{1}{3}}} 15x^2\,dx\,dy $
so then
$\displaystyle\int_0^8 \int_{{(y-4)}/2}^{y^{1/3}} 15x^2\,dx \,dy
=\int_0^8\left|5x^3\large\right|_{x={(y-4)}/2}^{x=y^{1/3}} \,dy$
so what values of y do we use 0 and 8?

 

[MarkFL]

Re: 15.2.51 double integral

Okay we have:

$$I=\int_0^8\int_{\frac{y-4}{2}}^{y^{\frac{1}{3}}} 15x^2\,dx\,dy$$

I would next write:

$$I=5\int_0^8 \left[x^3\right]_{\frac{y-4}{2}}^{y^{\frac{1}{3}}}\,dy=5\int_0^8 \left(y^{\frac{1}{3}}\right)^3-\left(\frac{y-4}{2}\right)^3\,dy=\frac{5}{8}\int_0^8 8y-(y-4)^3\,dy$$

Now you can either expand the cubed binomial in the integrand, or you can can integrate the two terms in the integrand separately. I would likely choose the latter. :)