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. (use symmetry to divide the region into two equal volumes.)
$z = 64 - x^{2} - y^{2}$
$z = 64 - 8x$
graph
graph description
$v = 2$
$\int_{0}^{\square} \int_{0}^{\square} (\square) d y d x$

Explanation:

Step1: Find the intersection of \( z = 64 - x^2 - y^2 \) and \( z = 64 - 8x \)

Set \( 64 - x^2 - y^2=64 - 8x \), simplify to \( x^2 - 8x + y^2 = 0 \), complete the square for \( x \): \( (x - 4)^2 + y^2 = 16 \). This is a circle with center \( (4,0) \) and radius \( 4 \). For the region of integration, we consider the projection on the \( xy \)-plane. By symmetry, we can focus on the part where \( y\geq0 \) and double it. First, find the limits for \( x \): the circle \( (x - 4)^2 + y^2 = 16 \) has \( x \) ranging from \( 0 \) to \( 8 \) (since center at \( x = 4 \), radius \( 4 \), so \( x\in[0,8] \)). For a given \( x \), solve \( (x - 4)^2 + y^2 = 16 \) for \( y \): \( y=\sqrt{16-(x - 4)^2} \). Also, the upper surface is \( z_1=64 - x^2 - y^2 \) and the lower surface is \( z_2=64 - 8x \), so the height of the solid is \( z_1 - z_2=(64 - x^2 - y^2)-(64 - 8x)=8x - x^2 - y^2 \).

Step2: Determine the limits of integration

• For \( x \): from \( 0 \) to \( 8 \) (but wait, actually, when we use symmetry, we can take \( x \) from \( 0 \) to \( 8 \), and \( y \) from \( 0 \) to \( \sqrt{16-(x - 4)^2} \), then multiply by 2. But the given integral has a factor of 2, so the inner integral for \( y \) is from \( 0 \) to \( \sqrt{16-(x - 4)^2} \), and \( x \) from \( 0 \) to \( 8 \)? Wait, no, let's re - check. The circle \( (x - 4)^2 + y^2 = 16 \), when \( y = 0 \), \( x=0 \) or \( x = 8 \). So \( x \) ranges from \( 0 \) to \( 8 \), and for each \( x \), \( y \) ranges from \( 0 \) to \( \sqrt{16-(x - 4)^2} \). The integrand is \( (64 - x^2 - y^2)-(64 - 8x)=8x - x^2 - y^2 \). And since we use symmetry (the solid is symmetric about \( y = 0 \)), we integrate from \( y = 0 \) to \( y=\sqrt{16-(x - 4)^2} \) and multiply by 2. So the limits for \( x \) are from \( 0 \) to \( 8 \)? Wait, no, the center is at \( x = 4 \), radius \( 4 \), so \( x\in[0,8] \). For the inner integral (with respect to \( y \)), when \( x \) is in \( [0,8] \), \( y \) goes from \( 0 \) to \( \sqrt{16-(x - 4)^2} \). The integrand is \( (64 - x^2 - y^2)-(64 - 8x)=8x - x^2 - y^2 \). And the factor of 2 comes from the symmetry over \( y = 0 \) (we integrate for \( y\geq0 \) and double it). So the integral is \( V = 2\int_{x = 0}^{8}\int_{y = 0}^{\sqrt{16-(x - 4)^2}}(8x - x^2 - y^2)dydx \). Wait, but let's check the limits again. Wait, the circle \( (x - 4)^2 + y^2=16 \) can be rewritten as \( y^2=-x^2 + 8x \), so \( y=\sqrt{8x - x^2} \) (since \( y\geq0 \) for the symmetric part). So the limits for \( x \): when \( y = 0 \), \( 8x - x^2=0\Rightarrow x = 0 \) or \( x = 8 \). So \( x\in[0,8] \), \( y\in[0,\sqrt{8x - x^2}] \), and the integrand is \( (64 - x^2 - y^2)-(64 - 8x)=8x - x^2 - y^2 \). And since we use the symmetry (the solid is symmetric about the \( x \) - axis, so the volume above \( y = 0 \) and below \( y = 0 \) is equal, so we calculate the volume for \( y\geq0 \) and multiply by 2. So the integral is \( V = 2\int_{0}^{8}\int_{0}^{\sqrt{8x - x^2}}(8x - x^2 - y^2)dydx \). Wait, \( \sqrt{16-(x - 4)^2}=\sqrt{16-(x^{2}-8x + 16)}=\sqrt{8x - x^{2}} \), yes. So the limits for \( x \) are from \( 0 \) to \( 8 \), for \( y \) from \( 0 \) to \( \sqrt{8x - x^2} \), and the integrand is \( 8x - x^2 - y^2 \).

Answer:

The double integral is \( V = 2\int_{0}^{8}\int_{0}^{\sqrt{8x - x^2}}(8x - x^2 - y^2)dydx \), so the first blank (outer integral limit for \( x \)) is \( 8 \), the second blank (inner integral limit for \( y \)) is \( \sqrt{8x - x^2} \), and the integrand (the part in the parentheses) is \( 8x - x^2 - y^2 \). So filling in the blanks: the outer integral limit \( \int_{0}^{\boldsymbol{8}} \), inner integral limit \( \int_{0}^{\boldsymbol{\sqrt{8x - x^2}}} \), and the integrand \( \boldsymbol{8x - x^2 - y^2} \).