Is This Integral Calculating Area or Volume?

[karush]

$$\iint_\limits{R}(3x^5-y^2\sin{y}+5)
\,dA$$
$$R=[(x,y)|x^2+y^2 \le 5]$$

​

 

[MarkFL]

Re: 232.q1.5e dbl trig int

First, let's plot $R$...

View attachment 7266

Let's try vertical strips:

$$V=\int_{-\sqrt{5}}^{\sqrt{5}}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}} 3x^5-y^2\sin(y)+5\,dy\,dx$$

One thing that makes our life a great deal easier is to recognize that the following term in the integrand:

$$y^2\sin(y)$$

is an odd function, and since the limits are symmetrical about the $y$-axis, it goes to zero, so we may state:

$$V=\int_{-\sqrt{5}}^{\sqrt{5}}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}} 3x^5+5\,dy\,dx$$

$$V=\int_{-\sqrt{5}}^{\sqrt{5}}3x^5+5\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\,dy\,dx$$

$$V=2\int_{-\sqrt{5}}^{\sqrt{5}} \sqrt{5-x^2}\left(3x^5+5\right)\,dx$$

Again, we may use the odd-function rule to simplify this to:

$$V=10\int_{-\sqrt{5}}^{\sqrt{5}} \sqrt{5-x^2}\,dx$$

Now, we may use the even function rule to state:

$$V=20\int_{0}^{\sqrt{5}} \sqrt{5-x^2}\,dx$$

I will let you take it from here. :)

Can you set this up using horizontal strips? How about polar coordinates?

 

[karush]

Re: 232.q1.5e dbl trig int

would this not be Area not Volume

$=25\pi$