<p>Consider the region R bounded on top by z=1-x and on the side by y=1-x^2 which lies in the first octant.</p>
<p>Find the volume of the region R. ( I know i have to triple integrate but i don’t know what the interval limits or integrand should be. I need to know how to set up the integration.)</p>
<p>Thank you so much for your time!! Thanks again!!!</p>
<p>Obviously you triple integrate to the infinite power, form a system of equations with z=1-x and y=1-x^2 thus solving for a relative solution of the variables x, z, and y, and finally apply some good ol’ right triangle trig converting all rectangular coordinates to polar.</p>
<p>Thanks for the advice. I tried sketching a graph out of the top of my head and was able to figure out the intervals kinda lol
0<z<1-x, 0<y<x^2, 0<x<1
you mentioned converting from rectangular to polar, if that is the case which equation is the right one to convert? z=1-x or y=1-x^2 ? I know how to convert but in the functions i need x and y in the same function right? should i just do :
1-x^2-y=0 and that should be what i convert to polar?</p>
<p>Yes, of course to the last question. Polar is necessary to find the volume of region R. Without it, 1-x^2-y=0 won’t be triple integrated correctly. You should derive infinity, apply the intervals, divide the appropriate graph into shapes so that you can calculate each one separately, add the results together, and voila, region R’s volume has become common knowledge and discover that E truly does equal MC^2</p>