Question

1. Use a double integral to find the volume under the plane z=2x+y and over the rectangle R={(x,y) : 8 ≤ x ≤ , 3 ≤ y ≤ 4}.

2. Use a double integral to find the volume of the solid enclosed by the surface z= x^2 and the planes x=0, x=6, y=6, y=0, and z=0.

3. Use double integration to find the volume of the solid bounded by the cylinder x^2 + y^2 = 9 and the planes z=0 and z= 3-x.

Answer 1

1.

Volume below the function $z=f(x,y)$ and above the region R is given by $\iint \limits _R f(x,y)dA$

$V= \iint \limits _R f(x,y)dA=\int\limits _8 ^{10}\int \limits _3^4(2x+y)dy \;dx=\intop\limits _8^{10} \big[2xy+\frac{1}{2}y^2 \big]_3^4 dx=$

$=\int \limits _8^{10} (8x+8)-(6x+\frac{9}{2}) dx=\int\limits _8^{10} (2x+\frac{7}{2})dx=\big[x^2+ \frac{7}{2}x\big]_8^{10}=$

$= (100+35)-(64+28)=43$

Answer: 43.

2.

Using the same formula for $z=x^2, R=\{(x,y):0 \leq x \leq 6, 0 \leq y \leq 6\}$

$V= \iint \limits _R x^2dA=\int\limits _0 ^{6}\int \limits _0^6 x^2dy \;dx=\intop\limits _0^{6} \big[x^2y\big]_0^6 dx= \intop\limits _0^{6} 6x^2 dx= \big[2x^3\big]_0^6= 432$

Answer: 432.

3.

Using the same formula for $z=3-x, R=\{(x,y):x^2+y^2=9\}$

$V= \iint \limits _R(3-x)dA=\int\limits _{-3} ^{3}\int \limits _{-\sqrt {9-y^2}}^{\sqrt {9-y^2}}(3-x) \;dx \;dy=\intop\limits _{-3}^{3} \big[3x-\frac{1}{2}x^2\big]_ {-\sqrt {9-y^2}} ^ {\sqrt {9-y^2}} dy=$

$\int \limits _{-3}^{3}6 {\sqrt {9-y^2}}dy= 6 * (\pi 3^2)/2=27\pi$