Sketch the region enclosed by $x+y^2=30$ and $x+y=0$. Decide whether to integrate with respect to $x$ or $y$. Then find the area of the region.
How do I start off sketching? Very first step?
I'm in Calc 2 and it's sad that I don't know what to do but I am trying to learn.
On
Try to identify the types of the functions you need to sketch by rewriting them in more familiar forms. For instance, you can write the first equation as $y^2=-(x-30)$. This is a parabola that has the same shape as $y^2 = x$. This latter function is a parabola with its axis of symmetry parallel to the $x$ axis, opening to the right, with its vertex at the origin. To understand what $y^2 = -(x-30)$ looks like, we need to reflect this parabola around the $y$ axis (so it will now open to the left; this is because of the negative sign in front). Then, we shift the parabola rightward by $30$ units. Thus, the curve is a parabola opening to the left, with its vertex at $(30, 0)$.
The second equation is clearly linear because it only involves single powers of $x$ and $y$. It can be rewritten as $y = -x$ which is a line with a $y$ intercept of $0$, and a slope $-1$.
Now you know the general shapes of each of the curves. To finish, you should determine where the curves intersect. This can be done by solving the system of equations using substitution. Knowing the general shapes of each curve, and knowing the points where they intersect should be enough to sketch the curves.
On
You have some good online sketchers to help you to visualize the functions and compare it with your results when you are a little bit lost. For instance in the case of writing the functions in terms of $y=f(x)$ (e.g in this page) this is how it looks like:
In blue: $y=\sqrt{30-x}$
In red: $y=-x$
And from that point you can continue to develop your solution to the problem.
And in terms of $x=f(y)$ (link here) it looks like this:
In blue: $x=30-y^2$
In red: $x=-y$
Can you sketch $y=30-x^2$? This is fairly standard highschool stuff. Once you do that, flip it over the line $y=x$ to interchange the $x$ and $y$ coordinates, to get $x=30-y^2$. That’s your first graph. The other one is easier, it’s $y=-x$ (you can also think of it as $x=-y$), still standard highschool stuff. With your two pictures, you are ready to go. You’ll need to know the points of intersection, those are where $x=30-y^2$ and $x=-y$, in other words where $30-y^2=-y$. Solve for $y$ by putting them all on one side of the equals sign, $0$ on the other side, and then factoring your quadratic in $y$.