Question 1

Evaluate the double integral  for the given function f(x, y) and the region R. f(x, y) = 5  ; R is bounded by the lines x = 1, y = 0 and y = x.
A) 
B) 
C) 
D) 
E)

Accepted Answer

A

Question 2

Use a double integral to find the volume of the solid shown in the figure. 
A) 
B) 
C) 
D)

Accepted Answer

A

Question 3

Evaluate the double integral  for the given function f(x, y) and the region R. f(x, y) = 6x + 6y; R is bounded by x = 0,  , y = 0 and y = 4.
A) 
B) 
C) 
D)

Accepted Answer

B

Question 4

Find the volume of the solid bounded above by the surface z = f(x, y)
And below by the plane region R.  ; R is the triangle with vertices (0, 0), (5, 0) and (0, 5).
A) 
B) 
C)

Accepted Answer

The answer of Find the volume of the solid bounded...

Question 5

Evaluate the double integral  for the given function f(x, y) and the region R.  ; R is bounded by  and y = x.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 6

Evaluate the double integral  for the given function f(x, y) and the region R.  R is the rectangle defined by 
A)10e
B)9
C)- 9e
D)- 10

Accepted Answer

We need to evaluate the double integral of $f(x,y)$ over the rectangular region R. Given function $f(x, y) = xy\cdot e^{xy}$.
Hence, 
$$\iint_R f(x,y)dA = \int_{0}^{1}\int_{0}^{2} xy\cdot e^{xy} \, dx\, dy$$
Integrating w.r.t $x$ using integration by parts,
$$\int xy\cdot e^{xy} \, dx = \frac{x}{2}\cdot e^{xy} - \frac{1}{2}\cdot e^{xy}+C_1$$
Where $C_1$ is a constant of integration.
Therefore, we get 
$$=\int_{0}^{1}\left[\frac{x}{2}\cdot e^{xy} - \frac{1}{2}\cdot e^{xy}ight]_0^2dy$$
$$=\int_{0}^{1}\frac{1}{2}(2y\cdot e^{2y} - e^{2y} - 0\cdot e^0 + e^0) dy$$
$$=\int_{0}^{1}\left[y\cdot e^{2y} - \frac{1}{2}\cdot e^{2y} + \frac{1}{2}ight] dy$$
Integrating w.r.t $y$ again,
$$\int y\cdot e^{2y} dy = \frac{1}{2}\cdot e^{2y}(y-\frac{1}{2})+C_2$$
Where $C_2$ is another constant of integration.
Therefore, we get 
$$=\left[\frac{1}{2}\cdot e^{2y}(y-\frac{1}{2})- \frac{1}{4}\cdot e^{2y} + \frac{1}{2}yight]_0^1$$
$$=\frac{1}{2}\cdot e^2 - \frac{3}{4}$$
Hence the answer is $\boxed{	extbf{(B)}\ 9}$.

Question 7

Evaluate the double integral  for the given function f(x, y) and the region R.  ; R is the rectangle defined by 
A) 
B) 
C) 
D)

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 8

Evaluate the double integral  for the given function f(x, y) and the region R.  ; R is bounded by x = 0, x = 1, y = 0 and y = x.
A)4
B)2e
C)- 2
D)- 4

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 9

Use a double integral to find the volume of the solid shown in the figure. 
A)14
B)6
C)4
D)9

Accepted Answer

The answer of Use a double integral to find the...

Question 10

Evaluate the double integral  for the given function f(x, y) and the region R. f(x, y) = 4x + 2y; R is bounded by x = 0, x = 4, y = 0 and y = x.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 11

Evaluate the double integral  for the function  and the region  .  and  is the rectangle defined by  and  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 12

Evaluate the double integral  for the given function f(x, y) and the region R. f(x, y) = 6x + 12y; R is bounded by x = 1, x = 3, y = 0 and y = x + 1.
A) 
B) 
C) 
D)

Accepted Answer

To evaluate the integral, we need to first find the bounds of integration for x and y. 
From the given region, we can see that x varies from 1 to 3 and y varies from 0 to x + 1. 
Thus, we have the integral:

$\int_1^3\int_0^{x+1} (6x+12y) dydx$

Evaluating the integral, we get:

$\int_1^3\int_0^{x+1} (6x+12y) dydx = \int_1^3 [ 6xy + 6y^2 ]_0^{x+1} dx$

Simplifying, we get:

$\int_1^3 [ 6x(x+1) + 6(x+1)^2 ]dx = \frac{56}{3}$

Therefore, the answer is B.

Question 13

Evaluate the double integral  for the given function f(x, y) and the region R.  ; R is the rectangle defined by 
A)5
B)0
C)3
D)- 5

Accepted Answer

Since f(x,y) = 0, the integral of f(x,y) over any region R will always be 0. Therefore, the double integral of f(x,y) over the region R is equal to 0. Hence, the correct answer is B) 0.

Question 14

Evaluate the double integral  for the given function f(x, y) and the region R. 
A)14
B)13
C)11
D)18
E)2

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 15

Evaluate the double integral  for the function  and the region  .  and  is bounded by  ,  ,  and  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the double integral  for the...

Question 16

Find the volume of the solid bounded above by the surface z = f(x, y)
And below by the plane region R.  ; R is the region bounded by  .
A) 
B) 
C)

Accepted Answer

Using the formula for volume under a surface, we have:

Volume = ∬R f(x, y) dA

By converting to polar coordinates, we have:

Volume = ∫0^2π ∫0^3 16 - r^2 r dr dθ

Solving this integral, we get:

Volume = 32π/3

Therefore, the correct answer is B, which is 11eb12b0_a91d_844b_b394_231e7ef824c7_TB7866_11.

Question 17

Use a double integral to find the volume of the solid shown in the figure. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

Notice that the solid is bounded below by the xy-plane and bounded above by the surface $z=16-x^2-2y^2$. Thus, we can find the volume of the solid using the double integral $$\iint_D \left(16-x^2-2y^2ight)\,dA,$$ where $D$ is the region in the xy-plane bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$. Switching to polar coordinates, we have \begin{align*} \iint_D \left(16-x^2-2y^2ight)\,dA &= \int_{-\sqrt{2}}^\sqrt{2}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}}\left(16-x^2-2y^2ight)\,dx\,dy\ &= \int_0^{2\pi}\int_0^\sqrt{2}\left(16-r^2\cos^2	heta-2r^2\sin^2	hetaight)r\,dr\,d	heta\ &= \int_0^{2\pi}\int_0^\sqrt{2}\left(16r-r^3\cos^2	heta-2r^3\sin^2	hetaight)\,dr\,d	heta\ &= \int_0^{2\pi}\left[\frac{8r^2}{3}-\frac{r^4}{4}\cos^2	heta-\frac{r^4}{2}\sin^2	hetaight]_0^\sqrt{2}\,d	heta\ &= \int_0^{2\pi}\left(\frac{16}{3}-\frac{1}{2}\cos^2	heta-\sin^2	hetaight)\,d	heta\ &= \left[\frac{16	heta}{3}-\frac{	heta}{4}-\frac{\cos	heta\sin	heta}{2}+\frac{	heta}{2}ight]_0^{2\pi}\ &= \frac{32\pi}{3}-\pi\ &= \frac{29}{3}\pi. \end{align*} Therefore, the answer is (C).

Question 18

Evaluate the double integral  for the given function f(x, y) and the region R. f(x, y) = 2y + x; R is the rectangle defined by 
A) 
B) 
C) 
D)26

Accepted Answer

We evaluate the double integral using the formula:

∬R f(x,y) dA = ∫[a,b] ∫[c,d] f(x,y) dy dx

= ∫[-2,2] ∫[1,3] (2y+x) dy dx

= ∫[-2,2] (y^2+xy) [1,3] dx

= ∫[-2,2] (8+2x) dx

= [8x+x^2] [-2,2]

= 16

Therefore, the answer is C) 11eb12b0_a918_54cb_b394_7bcc4f667f05_TB7866_11.

Question 19

Use a double integral to find the volume of the solid shown in the figure. 
A)5,996
B)5,994
C)6,000
D)6,008

Accepted Answer

We can use the double integral to find the volume of the solid:

$$V = \iint_D f(x,y) \,dA$$

In this case, $f(x,y) = 4 - x^2 - y^2$, which is the height of the solid at each point $(x,y)$ in the region $D$:

$$D = \{(x,y) \mid x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$$

We can convert this to polar coordinates to simplify the integral:

$$\begin{aligned} V &= \iint_D (4 - x^2 - y^2) \,dA \ &= \int_0^{\pi/2} \int_0^2 (4 - r^2) r \,dr\,d	heta \ &= \int_0^{\pi/2} (2r^2 - \frac{1}{3}r^4) \Big|_{r=0}^{r=2} \,d	heta \ &= \int_0^{\pi/2} (8 - \frac{32}{3}) \,d	heta \ &= \frac{16}{3} \pi \end{aligned}$$

So the volume of the solid is approximately 6,000 cubic units, which means the answer is C.

Question 20

Evaluate the double integral  for the given function f(x, y) and the region R. 
A)9
B)4
C)72
D)63
E)8

Accepted Answer

First, we need to find the limits of integration.

For x: 0 ≤ x ≤ 3
For y: x ≤ y ≤ 3

Then we can evaluate the integral:

∫[0,3]∫[x,3] 4y dA
= ∫[0,3] (4y(y-x)) dx
= ∫[0,3] (4y^2 - 4xy) dx
= [4/3 * y^3 - 2xy^2] from y=x to y=3
= [(4/3)*27 - 18x] - [(4/3)*x^3 - 2x^3] 
= 72/3 - 2x^3
= 72/3 - 2(3^3)
= 72/3 - 54
= 18

Therefore, the answer is C.

Quiz 8: Calculus of Several Variables

Evaluate the double integral for the given function f(x, y) and the region R. f(x, y) = 5 ; R is bounded by the lines x = 1, y = 0 and y = x.

Use a double integral to find the volume of the solid shown in the figure.

Evaluate the double integral for the given function f(x, y) and the region R. f(x, y) = 6x + 6y; R is bounded by x = 0, , y = 0 and y = 4.

Find the volume of the solid bounded above by the surface z = f(x, y) And below by the plane region R. ; R is the triangle with vertices (0, 0) , (5, 0) and (0, 5) .

Evaluate the double integral for the given function f(x, y) and the region R. ; R is bounded by and y = x.

Evaluate the double integral for the given function f(x, y) and the region R. R is the rectangle defined by

Evaluate the double integral for the given function f(x, y) and the region R. ; R is the rectangle defined by

Evaluate the double integral for the given function f(x, y) and the region R. ; R is bounded by x = 0, x = 1, y = 0 and y = x.

Use a double integral to find the volume of the solid shown in the figure.

Evaluate the double integral for the given function f(x, y) and the region R. f(x, y) = 4x + 2y; R is bounded by x = 0, x = 4, y = 0 and y = x.

Evaluate the double integral for the function and the region . and is the rectangle defined by and .

Evaluate the double integral for the given function f(x, y) and the region R. f(x, y) = 6x + 12y; R is bounded by x = 1, x = 3, y = 0 and y = x + 1.

Evaluate the double integral for the given function f(x, y) and the region R. ; R is the rectangle defined by

Evaluate the double integral for the given function f(x, y) and the region R.

Evaluate the double integral for the function and the region . and is bounded by , , and .

Find the volume of the solid bounded above by the surface z = f(x, y) And below by the plane region R. ; R is the region bounded by .

Use a double integral to find the volume of the solid shown in the figure.

Evaluate the double integral for the given function f(x, y) and the region R. f(x, y) = 2y + x; R is the rectangle defined by

Use a double integral to find the volume of the solid shown in the figure.

Evaluate the double integral for the given function f(x, y) and the region R.