Answer to Question #100912 in Calculus for Yenco

Question #100912
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.
1
Expert's answer
2020-01-05T17:14:57-0500

1.

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

V=Rf(x,y)dA=81034(2x+y)dy  dx=810[2xy+12y2]34dx=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=

=810(8x+8)(6x+92)dx=810(2x+72)dx=[x2+72x]810==\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= (100+35)-(64+28)=43

Answer: 43.

2.

Using the same formula for z=x2,R={(x,y):0x6,0y6}z=x^2, R=\{(x,y):0 \leq x \leq 6, 0 \leq y \leq 6\}

V=Rx2dA=0606x2dy  dx=06[x2y]06dx=066x2dx=[2x3]06=432V= \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=3x,R={(x,y):x2+y2=9}z=3-x, R=\{(x,y):x^2+y^2=9\}

V=R(3x)dA=339y29y2(3x)  dx  dy=33[3x12x2]9y29y2dy=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=

3369y2dy=6(π32)/2=27π\int \limits _{-3}^{3}6 {\sqrt {9-y^2}}dy= 6 * (\pi 3^2)/2=27\pi




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment