Question

set up a double integral to find the volume of the solid region bounded by the graphs of the equations. do not evaluate the integral.
$v = 4\int_{0}^{}\int_{0}^{}(\\ )dydx$
$z = x^{2}+y^{2}$
$x^{2}+y^{2}=49$
$z = 0$

Explanation:

Step1: Determine the region in the xy - plane

The curves in the xy - plane are given by \(x^{2}+y^{2}=49\) (a circle with radius \(r = 7\)) and we also have the limits for \(x\) from \(0\) to \(4\) (from the outer integral \(\int_{0}^{4}\)). For a given \(x\) in the interval \([0,4]\), we need to find the limits for \(y\). From the circle equation \(x^{2}+y^{2}=49\), we can solve for \(y\): \(y=\pm\sqrt{49 - x^{2}}\). But since we are probably dealing with the region in the first quadrant (because the outer integral for \(x\) is from \(0\) to \(4\) and we can assume the inner integral for \(y\) is from \(0\) to \(\sqrt{49 - x^{2}}\) if we consider the region where \(y\geq0\))? Wait, no, the volume is bounded by \(z=x^{2}+y^{2}\) and \(z = 0\), and also \(x^{2}+y^{2}=49\). Wait, the outer integral for \(x\) is from \(0\) to \(4\), so for each \(x\in[0,4]\), \(y\) ranges from \(0\) to \(\sqrt{49 - x^{2}}\)? Wait, no, the problem has an outer integral \(\int_{0}^{4}\) and an inner integral \(\int_{0}^{\square}\), so \(y\) ranges from \(0\) to \(\sqrt{49 - x^{2}}\)? Wait, no, the limits for \(x\) are from \(0\) to \(4\), and for \(y\), since the circle is \(x^{2}+y^{2}=49\), when \(x\in[0,4]\), \(y\) ranges from \(0\) to \(\sqrt{49 - x^{2}}\) (assuming we are in the first quadrant for \(y\) as well, because the inner integral starts from \(0\)).

Step2: Determine the integrand

The volume between two surfaces \(z = f(x,y)\) and \(z = g(x,y)\) is given by the double integral \(\iint_{R}(f(x,y)-g(x,y))dA\). Here, \(f(x,y)=x^{2}+y^{2}\) and \(g(x,y)=0\), so the integrand is \((x^{2}+y^{2})-0=x^{2}+y^{2}\).

Step3: Set up the double integral

The outer integral is with respect to \(x\) from \(0\) to \(4\), the inner integral is with respect to \(y\) from \(0\) to \(\sqrt{49 - x^{2}}\), and the integrand is \(x^{2}+y^{2}\). So the double integral for the volume \(V\) is:

\(V=\int_{0}^{4}\int_{0}^{\sqrt{49 - x^{2}}}(x^{2}+y^{2})dydx\)

Answer:

The double integral is \(\int_{0}^{4}\int_{0}^{\sqrt{49 - x^{2}}}(x^{2}+y^{2})dydx\). So the first blank (outer integral limit for \(x\) is already \(4\), the inner integral limit for \(y\) is \(\sqrt{49 - x^{2}}\), and the integrand is \(x^{2}+y^{2}\)). So filling in the blanks: the inner integral limit for \(y\) is \(\sqrt{49 - x^{2}}\), and the integrand is \(x^{2}+y^{2}\). So the double integral is \(\int_{0}^{4}\int_{0}^{\sqrt{49 - x^{2}}}(x^{2}+y^{2})dydx\)