Double integrals volume

Question

Find the volume below $z = 5+3y$ above the region $−5 \leqslant x \leqslant 5$, $0 \leqslant y \leqslant 25−x^2$.

How do I solve this? I don't know how to make equation to solve this problem. Anyone help?

Answer 1

If you want to integrate over a region $\Omega$ in the $x-y$ plane to find the volume of the solid with height $z$ and base $\Omega$, you can set up a double integral $$ \iint_{\Omega} f(x,y) \text{ d}A \ = \int_{-5}^{5} \int_{0}^{25-x^2} f(x,y) \text{ d}y \text{ d}x $$ where $f(x,y) = z = 5+3y$ is the height of the solid.

Answer 2

Our bounds on $x$ are $[-5,5]$ and our bounds of $y = [0, 25-x^2]$. We know that the volume is actually going to be a number and not in terms of $x$. So that means that our outer integral must have the bounds $[-5,5]$ and the inner must have the bounds $[0, 25-x^2]$. So now we have: $$\int_{-5}^{5} \int_0^{25-x^2} 5+3y \; \mathrm{d}y \;\mathrm{d}x $$ Which is easy to calculate.

Answer 3

Set it up like this:

$\begin{align} V &= \int_{-5}^5 \int_0^{25-x^2} f(x,y) \hspace{.1cm} dy \hspace{.05cm} dx \\ &= \int_{-5}^5 \int_0^{25-x^2} (5+3y) \hspace{.1cm} dy \hspace{.05cm} dx \end{align} $

The antiderivatives should be pretty simple, so you can take it from here.